Let Us Work Out 1.1
1. Which of the following polynomial is a quadratic polynomial :
(i) x² − 7x + 2
(ii) 7x⁵ − x(x + 2)
(iii) 2x(x + 5) + 1
(iv) 2x − 1
Solution :
A polynomial whose highest power of the variable is 2 is called a quadratic polynomial.
(i) x² − 7x + 2
Highest power of x = 2
∴ It is a quadratic polynomial.
(ii) 7x⁵ − x(x + 2)
= 7x⁵ − x² − 2x
Highest power of x = 5
∴ It is not a quadratic polynomial.
(iii) 2x(x + 5) + 1
= 2x² + 10x + 1
Highest power of x = 2
∴ It is a quadratic polynomial.
(iv) 2x − 1
Highest power of x = 1
∴ It is not a quadratic polynomial.
∴ Required quadratic polynomials are :
(i) x² − 7x + 2 and (iii) 2x(x + 5) + 1
2. Which of the following equations can be written in the form ax² + bx + c = 0 where a ≠ 0 ?
(i) x − 1 +
1
x
= 6
(ii) x +
3
x
= x²
(iii) x² − 6x + 2 = 0
(iv) (x − 2)² = x² − 4x + 4
Solution :
(i)
x − 1 +
1
x
= 6
Taking LCM = x
x² − x + 1
x
=
6x
x
x² − x + 1 = 6x
x² − 7x + 1 = 0
It is in the form ax² + bx + c = 0
∴ It is a quadratic equation.
(ii)
x +
3
x
= x²
Taking LCM = x
x² + 3
x
=
x³
x
x² + 3 = x³
x³ − x² − 3 = 0
Highest power of x = 3
∴ It is not a quadratic equation.
(iii)
x² − 6x + 2 = 0
It is already in the form ax² + bx + c = 0
∴ It is a quadratic equation.
(iv)
(x − 2)² = x² − 4x + 4
x² − 4x + 4 = x² − 4x + 4
0 = 0
No quadratic term remains.
∴ It is not a quadratic equation.
3. Let us determine the power of the variable in which the equation x⁶ − x³ − 2 = 0 becomes a quadratic equation.
Solution :
Let,
x³ = y
Then,
x⁶ = (x³)²
x⁶ = y²
Substituting in the given equation,
x⁶ − x³ − 2 = 0
y² − y − 2 = 0
Highest power of y = 2
∴ The equation becomes quadratic in x³.
4(i). Determine the value of a for which the equation (a − 2)x² + 3x + 5 = 0 will not be a quadratic equation.
Solution :
For a quadratic equation,
coefficient of x² ≠ 0
a − 2 ≠ 0
For the equation not to be quadratic,
a − 2 = 0
a = 2
∴ The equation will not be quadratic when a = 2.
4(ii). If
x
4 − x
=
1
3x
express it in the form ax² + bx + c = 0 and determine the coefficient of x.
Solution :
x
4 − x
=
1
3x
Taking LCM = 3x(4 − x)
3x × x = 4 − x
3x² = 4 − x
3x² + x − 4 = 0
Comparing with ax² + bx + c = 0
a = 3 , b = 1 , c = −4
∴ Coefficient of x = 1
4(iii). Express 3x² + 7x + 23 = (x + 4)(x + 3) + 2 in the form ax² + bx + c = 0.
Solution :
3x² + 7x + 23 = (x + 4)(x + 3) + 2
3x² + 7x + 23 = x² + 3x + 4x + 12 + 2
3x² + 7x + 23 = x² + 7x + 14
3x² + 7x + 23 − x² − 7x − 14 = 0
2x² + 9 = 0
∴ Required quadratic equation is :
2x² + 9 = 0
4(iv). Express the equation (x + 2)³ = x(x² − 1) in the form ax² + bx + c = 0 and write the coefficients of x², x and constant term.
Solution :
(x + 2)³ = x(x² − 1)
x³ + 6x² + 12x + 8 = x³ − x
x³ + 6x² + 12x + 8 − x³ + x = 0
6x² + 13x + 8 = 0
Comparing with ax² + bx + c = 0
a = 6 , b = 13 , c = 8
∴ Coefficient of x² = 6
∴ Coefficient of x = 13
∴ Constant term = 8
5(i). Divide 42 into two parts such that one part is equal to the square of the other part.
Solution :
Let one part = x
Then other part = x²
According to the question,
x + x² = 42
x² + x − 42 = 0
∴ Required quadratic equation is :
x² + x − 42 = 0
5(ii). The product of two consecutive positive odd numbers is 143.
Solution :
Let first odd number = x
Then second consecutive odd number = x + 2
According to the question,
x(x + 2) = 143
x² + 2x = 143
x² + 2x − 143 = 0
∴ Required quadratic equation is :
x² + 2x − 143 = 0
5(iii). The sum of the squares of two consecutive numbers is 313.
Solution :
Let first number = x
Then second consecutive number = x + 1
According to the question,
x² + (x + 1)² = 313
x² + x² + 2x + 1 = 313
2x² + 2x + 1 − 313 = 0
2x² + 2x − 312 = 0
x² + x − 156 = 0
∴ Required quadratic equation is :
x² + x − 156 = 0
6(i). The length of the diagonal of a rectangular area is 15 m and the length exceeds its breadth by 3 m.
Solution :
Let breadth of the rectangle = x m
Then length of the rectangle = (x + 3) m
Diagonal of the rectangle = 15 m
By Pythagoras theorem,
(Diagonal)² = (Length)² + (Breadth)²
15² = (x + 3)² + x²
225 = x² + 6x + 9 + x²
225 = 2x² + 6x + 9
2x² + 6x + 9 − 225 = 0
2x² + 6x − 216 = 0
x² + 3x − 108 = 0
∴ Required quadratic equation is :
x² + 3x − 108 = 0
6(ii). One person bought some sugar at ₹80. If he would get 4 kg more sugar with that money, then the price of 1 kg sugar would be less by ₹1.
Solution :
Let present price of sugar = ₹x per kg
Quantity of sugar purchased =
80
x
kg
If the price becomes ₹1 less,
New price = ₹(x − 1) per kg
Then quantity purchased =
80
x − 1
kg
According to the question,
80
x − 1
=
80
x
+ 4
Taking LCM = x(x − 1)
80x = 80(x − 1) + 4x(x − 1)
80x = 80x − 80 + 4x² − 4x
4x² − 4x − 80 = 0
x² − x − 20 = 0
∴ Required quadratic equation is :
x² − x − 20 = 0
6(iii). The distance between two stations is 300 km. A train went from first station to second station with uniform velocity. If the velocity of the train could be 5 km/hour more, then the time taken would be lesser by 2 hours.
Solution :
Let velocity of the train = x km/hour
Time taken to travel 300 km =
300
x
hours
If velocity increases by 5 km/hour,
New velocity = (x + 5) km/hour
New time taken =
300
x + 5
hours
According to the question,
300
x
−
300
x + 5
= 2
Taking LCM = x(x + 5)
300(x + 5) − 300x = 2x(x + 5)
300x + 1500 − 300x = 2x² + 10x
1500 = 2x² + 10x
2x² + 10x − 1500 = 0
x² + 5x − 750 = 0
∴ Required quadratic equation is :
x² + 5x − 750 = 0
6(iv). A clock seller sold a clock by purchasing it at ₹336. The amount of his profit percentage is as much as the amount with which he bought the clock.
Solution :
Cost price of the clock = ₹336
Let profit percentage = x%
Profit amount =
336 × x
100
According to the question,
profit amount = x
336x
100
= x
Taking LCM = 100
336x = 100x
236x = 0
∴ Required equation is :
236x = 0
6(v). If the velocity of the stream is 2 km/hr, then the total time taken by Ratanmajhi to cover 21 km downstream and the same distance upstream is 10 hours.
Solution :
Let velocity of the boat in still water = x km/hr
Velocity downstream = (x + 2) km/hr
Velocity upstream = (x − 2) km/hr
Time taken downstream =
21
x + 2
hours
Time taken upstream =
21
x − 2
hours
According to the question,
21
x + 2
+
21
x − 2
= 10
Taking LCM = (x + 2)(x − 2)
21(x − 2) + 21(x + 2) = 10(x² − 4)
21x − 42 + 21x + 42 = 10x² − 40
42x = 10x² − 40
10x² − 42x − 40 = 0
5x² − 21x − 20 = 0
∴ Required quadratic equation is :
5x² − 21x − 20 = 0
6(vi). The time taken to clean a garden by Majid is 3 hours more than Mahim. Together they complete the work in 2 hours.
Solution :
Let time taken by Mahim = x hours
Then time taken by Majid = (x + 3) hours
Work done by Mahim in 1 hour =
1
x
Work done by Majid in 1 hour =
1
x + 3
Together they complete the work in 2 hours.
1
x
+
1
x + 3
=
1
2
Taking LCM = 2x(x + 3)
2(x + 3) + 2x = x(x + 3)
2x + 6 + 2x = x² + 3x
4x + 6 = x² + 3x
x² − x − 6 = 0
∴ Required quadratic equation is :
x² − x − 6 = 0
6(vii). The unit digit of a two digit number exceeds its tens digit by 6 and the product of the digits is less than the number by 12.
Solution :
Let tens digit = x
Then unit digit = x + 6
The number =
10x + (x + 6)
= 11x + 6
Product of the digits =
x(x + 6)
According to the question,
x(x + 6) = (11x + 6) − 12
x² + 6x = 11x − 6
x² − 5x + 6 = 0
∴ Required quadratic equation is :
x² − 5x + 6 = 0
6(viii). There is a road of equal width around a rectangular playground outside it having length 45 m and breadth 40 m. Area of the road is 450 sq.m.
Solution :
Let width of the road = x m
Outer length = (45 + 2x) m
Outer breadth = (40 + 2x) m
Outer area =
(45 + 2x)(40 + 2x)
Area of the playground =
45 × 40 = 1800 sq.m
Area of the road = 450 sq.m
(45 + 2x)(40 + 2x) − 1800 = 450
1800 + 90x + 80x + 4x² − 1800 = 450
4x² + 170x = 450
4x² + 170x − 450 = 0
2x² + 85x − 225 = 0
∴ Required quadratic equation is :
2x² + 85x − 225 = 0
Let Us Work Out 1.2
1. In each of the following cases, let us justify and write whether the given values are the roots of the given quadratic equation :
(i) x² + x + 1 = 0 ; 1 and −1
(ii) 8x² + 7x = 0 ; 0 and −2
(iii) x +
1
x
=
13
6
;
5
6
and
4
3
(iv) x² − √3x − 6 = 0 ; −√3 and 2√3
Solution :
(i) For x = 1
x² + x + 1
= (1)² + 1 + 1
= 1 + 1 + 1
= 3 ≠ 0
∴ 1 is not a root of the equation.
Now, for x = −1
x² + x + 1
= (−1)² + (−1) + 1
= 1 − 1 + 1
= 1 ≠ 0
∴ −1 is not a root of the equation.
∴ 1 and −1 are not the roots of the equation.
(ii) For x = 0
8x² + 7x
= 8(0)² + 7(0)
= 0
∴ 0 is a root of the equation.
Now, for x = −2
8x² + 7x
= 8(−2)² + 7(−2)
= 8 × 4 − 14
= 32 − 14
= 18 ≠ 0
∴ −2 is not a root of the equation.
∴ Only 0 is a root of the equation.
(iii) For x =
5
6
x +
1
x
=
5
6
+
6
5
Taking LCM = 30
=
25 + 36
30
=
61
30
But,
13
6
=
65
30
Therefore,
61
30
≠
65
30
∴
5
6
is not a root.
Now, for x =
4
3
x +
1
x
=
4
3
+
3
4
Taking LCM = 12
=
16 + 9
12
=
25
12
But,
13
6
=
26
12
Therefore,
25
12
≠
26
12
∴
4
3
is not a root.
∴ None of the given values are roots of the equation.
(iv) For x = −√3
x² − √3x − 6
= (−√3)² − √3(−√3) − 6
= 3 + 3 − 6
= 0
∴ −√3 is a root of the equation.
Now, for x = 2√3
x² − √3x − 6
= (2√3)² − √3(2√3) − 6
= 12 − 6 − 6
= 0
∴ 2√3 is a root of the equation.
∴ −√3 and 2√3 are the roots of the equation.
2(i). Let us calculate and write the value of k for which
2
3
will be a root of the quadratic equation 7x² + kx − 3 = 0.
Solution :
Since
2
3
is a root of the equation,
7x² + kx − 3 = 0
Putting
x =
2
3
7
(
2
3
)²
+
k
(
2
3
)
− 3 = 0
7 ×
4
9
+
2k
3
− 3 = 0
28
9
+
2k
3
− 3 = 0
Taking LCM = 9
28 + 6k − 27
9
= 0
1 + 6k = 0
6k = −1
k =
−
1
6
∴ Required value of k =
−
1
6
2(ii). Let us calculate and write the value of k for which −a will be a root of the quadratic equation x² + 3ax + k = 0.
Solution :
Since −a is a root of the equation,
x² + 3ax + k = 0
Putting x = −a
(−a)² + 3a(−a) + k = 0
a² − 3a² + k = 0
−2a² + k = 0
k = 2a²
∴ Required value of k = 2a²
3. If
2
3
and −3 are the two roots of the quadratic equation ax² + 7x + b = 0, let us calculate the values of a and b.
Solution :
Since
2
3
and −3 are roots of the equation,
ax² + 7x + b = 0
Putting
x =
2
3
a
(
2
3
)²
+ 7
(
2
3
)
+ b = 0
a ×
4
9
+
14
3
+ b = 0
Taking LCM = 9
4a + 42 + 9b
9
= 0
4a + 9b + 42 = 0
4a + 9b = −42
Again, putting x = −3
a(−3)² + 7(−3) + b = 0
9a − 21 + b = 0
9a + b = 21
From,
9a + b = 21
b = 21 − 9a
Substituting in
4a + 9b = −42
4a + 9(21 − 9a) = −42
4a + 189 − 81a = −42
−77a = −231
a = 3
Now,
b = 21 − 9a
= 21 − 9 × 3
= 21 − 27
= −6
∴ Required values are :
a = 3 and b = −6
4(i). Solve : 3y² − 20 = 160 − 2y²
Solution :
3y² − 20 = 160 − 2y²
3y² + 2y² = 160 + 20
5y² = 180
y² = 36
∴ y = ±6
4(ii). Solve : (2x + 1)² + (x + 1)² = 6x + 47
Solution :
(2x + 1)² + (x + 1)² = 6x + 47
4x² + 4x + 1 + x² + 2x + 1 = 6x + 47
5x² + 6x + 2 = 6x + 47
5x² − 45 = 0
x² − 9 = 0
(x − 3)(x + 3) = 0
∴ x = 3 or x = −3
4(iii). Solve : (x − 7)(x − 9) = 195
Solution :
(x − 7)(x − 9) = 195
x² − 9x − 7x + 63 = 195
x² − 16x + 63 = 195
x² − 16x − 132 = 0
x² − 22x + 6x − 132 = 0
x(x − 22) + 6(x − 22) = 0
(x − 22)(x + 6) = 0
∴ x = 22 or x = −6
4(iv). Solve :
3x −
24
x
=
x
3
Solution :
Taking LCM = 3x
9x² − 72 = x²
8x² − 72 = 0
x² − 9 = 0
(x − 3)(x + 3) = 0
∴ x = 3 or x = −3
4(v). Solve :
x
3
+
3
x
=
15
x
Solution :
Taking LCM = 3x
x² + 9 = 45
x² − 36 = 0
(x − 6)(x + 6) = 0
∴ x = 6 or x = −6
4(vi). Solve :
10x −
1
x
= 3
Solution :
Taking LCM = x
10x² − 1 = 3x
10x² − 3x − 1 = 0
10x² − 5x + 2x − 1 = 0
5x(2x − 1) + 1(2x − 1) = 0
(2x − 1)(5x + 1) = 0
∴ x =
1
2
or
x =
−
1
5
4(viii). Solve :
(x − 2)
(x + 2)
+
6
(x − 2)
(x − 6)
= 1
Solution :
(x − 2)
(x + 2)
+
6
(x − 2)
(x − 6)
= 1
(x − 2)(x − 6) + 6(x − 2)(x + 2) = (x + 2)(x − 6)
x² − 8x + 12 + 6x² − 24 = x² − 4x − 12
7x² − 8x − 12 = x² − 4x − 12
6x² − 4x = 0
2x(3x − 2) = 0
∴ x = 0 or
x =
2
3
4(ix). Solve :
1
(x − 3)
−
1
(x + 5)
=
1
6
Solution :
6(x + 5) − 6(x − 3) = (x − 3)(x + 5)
6x + 30 − 6x + 18 = x² + 2x − 15
48 = x² + 2x − 15
x² + 2x − 63 = 0
x² + 9x − 7x − 63 = 0
x(x + 9) − 7(x + 9) = 0
(x + 9)(x − 7) = 0
∴ x = −9 or x = 7
4(x). Solve :
x
(x + 1)
+
(x + 1)
x
= 2
1
12
Solution :
12x² + 12(x + 1)² = 25x(x + 1)
12x² + 12x² + 24x + 12 = 25x² + 25x
x² + x − 12 = 0
x² + 4x − 3x − 12 = 0
x(x + 4) − 3(x + 4) = 0
(x + 4)(x − 3) = 0
∴ x = −4 or x = 3
4(xi). Solve :
ax + b
a + bx
=
cx + d
c + dx
Solution :
(ax + b)(c + dx) = (cx + d)(a + bx)
acx + adx² + bc + bdx = acx + bcx² + ad + bdx
adx² + bc = bcx² + ad
adx² − bcx² + bc − ad = 0
x²(ad − bc) − (ad − bc) = 0
(ad − bc)(x² − 1) = 0
(x − 1)(x + 1) = 0
∴ x = 1 or x = −1
4(xii). Solve :
(2x + 1) +
3
(2x + 1)
= 4
Solution :
(2x + 1)² + 3 = 4(2x + 1)
4x² + 4x + 1 + 3 = 8x + 4
4x² − 4x = 0
4x(x − 1) = 0
∴ x = 0 or x = 1
4(xiii). Solve :
(x + 1)
2
+
2
(x + 1)
=
(x + 1)
3
+
3
(x + 1)
−
5
6
Solution :
3(x + 1)² + 12 = 2(x + 1)² + 18 − 5(x + 1)
3x² + 6x + 3 + 12 = 2x² + 4x + 2 + 18 − 5x − 5
3x² + 6x + 15 = 2x² − x + 15
x² + 7x = 0
x(x + 7) = 0
∴ x = 0 or x = −7
4(xiv). Solve :
(12x + 17)
(3x + 1)
−
(2x + 15)
(x + 7)
= 3
1
5
Solution :
5(12x + 17)(x + 7) − 5(2x + 15)(3x + 1)
=
16(3x + 1)(x + 7)
60x² + 505x + 595 − 30x² − 235x − 75
=
48x² + 352x + 112
30x² + 270x + 520 = 48x² + 352x + 112
18x² + 82x − 408 = 0
9x² + 41x − 204 = 0
9x² + 68x − 27x − 204 = 0
x(9x + 68) − 3(9x + 68) = 0
(9x + 68)(x − 3) = 0
∴ x = 3 or
x =
−
68
9
4(xv). Solve :
(x + 3)
(x − 3)
+
6
(x − 3)
(x + 3)
= 5
Solution :
(x + 3)² + 6(x − 3)² = 5(x − 3)(x + 3)
x² + 6x + 9 + 6x² − 36x + 54
=
5x² − 45
7x² − 30x + 63 = 5x² − 45
2x² − 30x + 108 = 0
x² − 15x + 54 = 0
x² − 9x − 6x + 54 = 0
x(x − 9) − 6(x − 9) = 0
(x − 9)(x − 6) = 0
∴ x = 9 or x = 6
4(xvi). Solve :
1
(a + b + x)
=
1
a
+
1
b
+
1
x
Solution :
abx
=
bx(a + b + x)
+
ax(a + b + x)
+
ab(a + b + x)
abx
=
abx + b²x + bx² + a²x + abx + ax² + a²b + ab² + abx
0
=
(a + b)(x² + ax + bx + ab)
0
=
(a + b)(x + a)(x + b)
∴ x = −a or x = −b
4(xvii). Solve :
(
(x + a)
(x − a)
)²
− 5
(x + a)
(x − a)
+ 6 = 0
Solution :
Let
(x + a)
(x − a)
= y
y² − 5y + 6 = 0
y² − 3y − 2y + 6 = 0
y(y − 3) − 2(y − 3) = 0
(y − 3)(y − 2) = 0
∴ y = 3 or y = 2
When y = 3
(x + a)
(x − a)
= 3
x + a = 3x − 3a
2x = 4a
x = 2a
When y = 2
(x + a)
(x − a)
= 2
x + a = 2x − 2a
x = 3a
∴ x = 2a or x = 3a
4(xviii). Solve :
1
x
−
1
(x + b)
=
1
a
−
1
(a + b)
Solution :
a(a + b)(x + b)
−
a(a + b)x
=
x(x + b)(a + b)
−
ax(x + b)
a²x + a²b + abx + ab² − a²x − abx
=
ax² + abx + bx² + b²x − ax² − abx
a²b + ab²
=
bx² + b²x
ab(a + b)
=
bx(x + b)
a(a + b)
=
x(x + b)
x² + bx − a² − ab = 0
x² + (a + b)x − ax − a(a + b) = 0
x(x + a + b) − a(x + a + b) = 0
(x − a)(x + a + b) = 0
∴ x = a or x = −(a + b)
4(xix). Solve :
1
(x − 1)(x − 2)
+
1
(x − 2)(x − 3)
+
1
(x − 3)(x − 4)
=
1
6
Solution :
6(x − 3)(x − 4)
+
6(x − 1)(x − 4)
+
6(x − 1)(x − 2)
=
(x − 1)(x − 2)(x − 3)(x − 4)
6(x² − 7x + 12)
+
6(x² − 5x + 4)
+
6(x² − 3x + 2)
=
(x² − 5x + 4)(x² − 5x + 6)
18x² − 90x + 108
=
x⁴ − 10x³ + 35x² − 50x + 24
x⁴ − 10x³ + 17x² + 40x − 84 = 0
(x² − 5x + 6)(x² − 5x − 14) = 0
(x − 2)(x − 3)(x² − 5x − 14) = 0
∴ x = 2 or x = 3 or
x =
5 + √81
2
or
x =
5 − √81
2
∴ x = 2, 3, 7, −2
4(xx). Solve :
a
(x − a)
+
b
(x − b)
=
2c
(x − c)
Solution :
a(x − b)(x − c)
+
b(x − a)(x − c)
=
2c(x − a)(x − b)
ax² − abx − acx + abc
+
bx² − abx − bcx + abc
=
2cx² − 2acx − 2bcx + 2abc
(a + b)x² − 2abx − (a + b)cx + 2abc
=
2cx² − 2acx − 2bcx + 2abc
(a + b)x² − 2abx − cx(a + b)
− 2cx² + 2cx(a + b)
= 0
(a + b − 2c)x²
+
(ca + cb − 2ab)x = 0
x[(a + b − 2c)x + ca + cb − 2ab] = 0
∴ x = 0
4(xxi). Solve :
x² − (√3 + 2)x + 2√3 = 0
Solution :
x² − (√3 + 2)x + 2√3 = 0
x² − √3x − 2x + 2√3 = 0
x(x − √3) − 2(x − √3) = 0
(x − √3)(x − 2) = 0
∴ x = √3 or x = 2
Let Us Work Out 1.3
1. The difference of two positive whole numbers is 3 and the sum of their squares is 117; by calculating, let us write the two numbers.
Solution :
Let the smaller number = x
Then the greater number = x + 3
x² + (x + 3)² = 117
x² + x² + 6x + 9 = 117
2x² + 6x − 108 = 0
x² + 3x − 54 = 0
x² + 9x − 6x − 54 = 0
x(x + 9) − 6(x + 9) = 0
(x + 9)(x − 6) = 0
x = −9 or x = 6
Since the numbers are positive,
x = 6
First number = 6
Second number = 9
2. The base of a triangle is 18 m more than two times of its height. If the area of the triangle is 360 sq.m., then let us determine the height of it.
Solution :
Let height of triangle = x m
Then base = (2x + 18) m
½ × x × (2x + 18) = 360
x(2x + 18) = 720
2x² + 18x − 720 = 0
x² + 9x − 360 = 0
x² + 24x − 15x − 360 = 0
x(x + 24) − 15(x + 24) = 0
(x + 24)(x − 15) = 0
x = −24 or x = 15
Since height cannot be negative,
x = 15
∴ Height of the triangle = 15 m
3. If 5 times of a positive whole number is less by 3 than twice of its square, then let us determine the number.
Solution :
Let the number = x
2x² − 3 = 5x
2x² − 5x − 3 = 0
2x² − 6x + x − 3 = 0
2x(x − 3) + 1(x − 3) = 0
(x − 3)(2x + 1) = 0
x = 3 or
x =
−
1
2
Since the number is a positive whole number,
x = 3
∴ Required number = 3
4. The distance between two places is 200 km. The time taken by a motor car is less by 2 hours than the time taken by a jeep car. If the speed of the motor car is 5 km/hr more than the speed of the jeep car, then let us write the speed of the motor car.
Solution :
Let speed of jeep = x km/hr
Then speed of motor car = (x + 5) km/hr
200
x
−
200
x + 5
= 2
200(x + 5) − 200x = 2x(x + 5)
1000 = 2x² + 10x
x² + 5x − 500 = 0
x² + 25x − 20x − 500 = 0
x(x + 25) − 20(x + 25) = 0
(x + 25)(x − 20) = 0
x = −25 or x = 20
Speed of jeep = 20 km/hr
Speed of motor car = 25 km/hr
5. The area of Amita's rectangular land is 2000 sq.m and perimeter is 180 m. By calculating, let us write the length and breadth of the land.
Solution :
Let breadth = x m
Then length = (90 − x) m
x(90 − x) = 2000
90x − x² = 2000
x² − 90x + 2000 = 0
x² − 50x − 40x + 2000 = 0
x(x − 50) − 40(x − 50) = 0
(x − 50)(x − 40) = 0
x = 50 or x = 40
∴ Length = 50 m and Breadth = 40 m
6. The tens digit of a two digit number is less by 3 than the units digit. If the product of the two digits is subtracted from the number, the result is 15. Let us write the unit digit of the number.
Solution :
Let tens digit = x
Then units digit = x + 3
Number = 10x + (x + 3)
= 11x + 3
Product of digits = x(x + 3)
(11x + 3) − x(x + 3) = 15
11x + 3 − x² − 3x = 15
−x² + 8x − 12 = 0
x² − 8x + 12 = 0
x² − 6x − 2x + 12 = 0
x(x − 6) − 2(x − 6) = 0
(x − 6)(x − 2) = 0
x = 6 or x = 2
Units digit = x + 3
= 9 or 5
7. Two pipes together take 11 minutes to fill a reservoir. One pipe takes 5 minutes more than the other pipe. Let us write the time taken separately by each pipe.
Solution :
Let first pipe take x minutes
Then second pipe takes (x + 5) minutes
1
x
+
1
x + 5
=
1
11
11(x + 5) + 11x = x(x + 5)
11x + 55 + 11x = x² + 5x
x² − 17x − 55 = 0
x² − 22x + 5x − 55 = 0
x(x − 22) + 5(x − 22) = 0
(x − 22)(x + 5) = 0
x = 22 or x = −5
∴ First pipe takes 22 minutes
∴ Second pipe takes 27 minutes
8. Porna and Pijush together complete a work in 4 days. If Porna takes 6 days more than Pijush, let us write the time taken by Porna alone.
Solution :
Let Pijush takes x days
Then Porna takes (x + 6) days
1
x
+
1
x + 6
=
1
4
4(x + 6) + 4x = x(x + 6)
4x + 24 + 4x = x² + 6x
x² − 2x − 24 = 0
x² − 6x + 4x − 24 = 0
x(x − 6) + 4(x − 6) = 0
(x − 6)(x + 4) = 0
x = 6 or x = −4
∴ Pijush takes 6 days
∴ Porna takes 12 days
9. If the price of 1 dozen pen is reduced by ₹6, then 3 more pens will be got for ₹30. Let us calculate the price of 1 dozen pen before reduction.
Solution :
Let original price of 1 dozen pens = ₹x
Number of pens bought for ₹30 =
360
x
New price of 1 dozen pens = ₹(x − 6)
New number of pens bought for ₹30 =
360
x − 6
360
x − 6
=
360
x
+ 3
360x = 360(x − 6) + 3x(x − 6)
360x = 360x − 2160 + 3x² − 18x
3x² − 18x − 2160 = 0
x² − 6x − 720 = 0
x² − 30x + 24x − 720 = 0
x(x − 30) + 24(x − 30) = 0
(x − 30)(x + 24) = 0
x = 30 or x = −24
∴ Original price of 1 dozen pens = ₹30
10. V.S.A.
(A) M.C.Q.
(i) The number of roots of a quadratic equation is :
(a) one
(b) two
(c) three
(d) none of them
Solution :
A quadratic equation has two roots.
∴ Correct answer = (b) two
(ii) 6x² + bx + c = 0 is a quadratic equation, then
(a) b ≠ 0
(b) c ≠ 0
(c) a ≠ 0
(d) none of these
Solution :
In a quadratic equation,
the coefficient of x² must not be zero.
Here coefficient of x² = 6
∴ Correct answer = (c) a ≠ 0
(iii) The highest power of the variable in a quadratic equation is :
(a) 1
(b) 2
(c) 3
(d) none of these
Solution :
The degree of a quadratic equation is 2.
∴ Highest power of the variable = 2
∴ Correct answer = (b) 2
(iv) The equation 4(5x² − 7x + 2) = 5(4x² − 6x + 3) is :
(a) linear
(b) quadratic
(c) 3rd degree
(d) none of these
Solution :
4(5x² − 7x + 2) = 5(4x² − 6x + 3)
20x² − 28x + 8 = 20x² − 30x + 15
20x² − 28x + 8 − 20x² + 30x − 15 = 0
2x − 7 = 0
Highest power of x = 1
∴ It is a linear equation.
∴ Correct answer = (a) linear
(v) The root / two roots of the equation
x²
x
= 6
(a) 0
(b) 6
(c) 0 and 6
(d) −6
Solution :
x²
x
= 6
x = 6
∴ Root of the equation = 6
∴ Correct answer = (b) 6
(B) Let us identify the following statements as True or False :
(i) (x − 3)² = x² − 6x + 9 is a quadratic equation.
Solution :
(x − 3)² = x² − 6x + 9
x² − 6x + 9 = x² − 6x + 9
0 = 0
It is an identity, not a quadratic equation.
∴ Statement is False.
(ii) 5 is the only root of the equation x² − 25 = 0
Solution :
x² − 25 = 0
(x − 5)(x + 5) = 0
x = 5 or x = −5
∴ 5 is not the only root.
∴ Statement is False.
(C) Let us fill in the blanks :
(i) If a = 0 and b ≠ 0 in the equation ax² + bx + c = 0, then the equation is a ______ equation.
Solution :
If a = 0, then ax² term disappears.
bx + c = 0
This is a linear equation.
∴ Blank = linear
(ii) If the two roots of a quadratic equation are 1 and −1, then the equation is ______
Solution :
(x − 1)(x + 1) = 0
x² − 1 = 0
∴ Required equation = x² − 1 = 0
(iii) The two roots of the equation x² = 6x are ______
Solution :
x² = 6x
x² − 6x = 0
x(x − 6) = 0
x = 0 or x = 6
∴ The two roots are 0 and 6.
11. S.A.
(i) Let us find the value of a, if one root of the equation x² + ax + 3 = 0 is 1.
Solution :
Since 1 is a root,
(1)² + a(1) + 3 = 0
1 + a + 3 = 0
a + 4 = 0
a = −4
∴ Required value of a = −4
(ii) Let us write the value of the other root if one root of the equation x² − (2 + b)x + 6 = 0 is 2.
Solution :
Let the other root be α
2 × Î± = 6
α = 3
∴ Other root = 3
(iii) Let us write the value of the other root if one root of the equation 2x² + kx + 4 = 0 is 2.
Solution :
Let the other root be α
2 × Î± =
4
2
2α = 2
α = 1
∴ Other root = 1
(iv) Let us write the equation if the difference of a proper fraction and its reciprocal is 50.
Solution :
Let the proper fraction =
x
1
Then reciprocal =
1
x
1
x
− x = 50
1 − x² = 50x
x² + 50x − 1 = 0
∴ Required equation = x² + 50x − 1 = 0
(v) Let us write the values of a and b, if the two roots of the equation ax² + bx + 35 = 0 are −5 and −7.
Solution :
ax² + bx + 35 = 0
Roots are −5 and −7
a(x + 5)(x + 7) = 0
a(x² + 12x + 35) = 0
ax² + 12ax + 35a = 0
Comparing with ax² + bx + 35 = 0
35a = 35
a = 1
b = 12
∴ a = 1 and b = 12
Let Us Work Out 1.4
1(i). Let us write whether Sridhara Acharyya's Formula is applicable to solve the equation
4x² + (2x − 1)(2x + 1) − 4x(2x − 1) = 0
Solution :
4x² + (4x² − 1) − (8x² − 4x) = 0
4x² + 4x² − 1 − 8x² + 4x = 0
4x − 1 = 0
The equation is linear.
∴ Sridhara Acharyya's Formula is not applicable.
1(ii). Let us write what type of equations can be solved with the help of Sridhara Acharyya's Formula.
Solution :
Sridhara Acharyya's Formula is used to solve quadratic equations.
ax² + bx + c = 0 , a ≠ 0
1(iii). By applying Sridhara Acharyya's Formula in the equation
5x² + 2x − 7 = 0
it is found that
x =
k ± 12
10
Let us calculate the value of k.
Solution :
Here,
a = 5 , b = 2 , c = −7
Using Sridhara Acharyya's Formula,
x =
−b ± √(b² − 4ac)
2a
x =
−2 ± √(2² − 4 × 5 × (−7))
10
x =
−2 ± √144
10
x =
−2 ± 12
10
∴ k = −2
2(i). Solve :
3x² + 11x − 4 = 0
Solution :
x =
−11 ± √(11² − 4 × 3 × (−4))
2 × 3
x =
−11 ± √169
6
x =
−11 ± 13
6
x =
2
6
or
x =
−24
6
x =
1
3
or x = −4
2(ii). Solve :
(x − 2)(x + 4) + 9 = 0
Solution :
x² + 4x − 2x − 8 + 9 = 0
x² + 2x + 1 = 0
(x + 1)² = 0
x = −1
2(iii). Solve :
(4x − 3)² − 2(x + 3) = 0
Solution :
16x² − 24x + 9 − 2x − 6 = 0
16x² − 26x + 3 = 0
x =
26 ± √(26² − 4 × 16 × 3)
32
x =
26 ± √484
32
x =
26 ± 22
32
x =
48
32
or
x =
4
32
x =
3
2
or
x =
1
8
2(iv). Solve :
3x² + 2x − 1 = 0
Solution :
x =
−2 ± √(2² − 4 × 3 × (−1))
6
x =
−2 ± √16
6
x =
−2 ± 4
6
x =
2
6
or
x =
−6
6
x =
1
3
or x = −1
2(v). Solve :
3x² + 2x + 1 = 0
Solution :
D = b² − 4ac
= 2² − 4 × 3 × 1
= 4 − 12
= −8
Since D < 0,
∴ No real roots.
2(vi). Solve :
10x² − x − 3 = 0
Solution :
x =
1 ± √(1 + 120)
20
x =
1 ± 11
20
x =
12
20
or
x =
−10
20
x =
3
5
or
x =
−
1
2
3(i). Sathi has drawn a right-angled triangle whose length of the hypotenuse is 6 cm more than twice of the shortest side. If the length of the third side is 2 cm less than the length of the hypotenuse, let us write the lengths of the three sides.
Solution :
Let shortest side = x cm
Hypotenuse = (2x + 6) cm
Third side = (2x + 4) cm
By Pythagoras theorem,
x² + (2x + 4)² = (2x + 6)²
x² + 4x² + 16x + 16 = 4x² + 24x + 36
x² − 8x − 20 = 0
x² − 10x + 2x − 20 = 0
x(x − 10) + 2(x − 10) = 0
(x − 10)(x + 2) = 0
x = 10 or x = −2
Since length cannot be negative,
x = 10
Shortest side = 10 cm
Third side = 24 cm
Hypotenuse = 26 cm
3(ii). If a two digit positive number is multiplied by its unit digit, then the product is 189 and if the tens digit is twice the unit digit, then let us calculate the unit digit.
Solution :
Let unit digit = x
Tens digit = 2x
Required number = 20x + x
= 21x
According to the question,
21x × x = 189
21x² = 189
x² = 9
x = ±3
Since digit cannot be negative,
x = 3
∴ Unit digit = 3
3(iii). The speed of Salma is 1 m/sec more than the speed of Anik. In a 180 m race, Salma reaches 2 seconds before than Anik. Let us write the speed of Anik.
Solution :
Let speed of Anik = x m/sec
Speed of Salma = (x + 1) m/sec
Time taken by Anik =
180
x
sec
Time taken by Salma =
180
x + 1
sec
According to the question,
180
x
−
180
x + 1
= 2
180(x + 1) − 180x = 2x(x + 1)
180 = 2x² + 2x
x² + x − 90 = 0
x² + 10x − 9x − 90 = 0
x(x + 10) − 9(x + 10) = 0
(x + 10)(x − 9) = 0
x = −10 or x = 9
Since speed cannot be negative,
x = 9
∴ Speed of Anik = 9 m/sec
3(iv). There is a square park in our locality. The area of a rectangular park is 78 sq.m less than twice of the area of that square-shaped park whose length is 5 m more than the side of the square and breadth is 3 m less than the side of the square. Let us write the length of the side of the square-shaped park.
Solution :
Let side of square park = x m
Area of square park = x²
Length of rectangle = (x + 5) m
Breadth of rectangle = (x − 3) m
Area of rectangle = (x + 5)(x − 3)
According to the question,
(x + 5)(x − 3) = 2x² − 78
x² + 2x − 15 = 2x² − 78
x² − 2x − 63 = 0
x² − 9x + 7x − 63 = 0
x(x − 9) + 7(x − 9) = 0
(x − 9)(x + 7) = 0
x = 9 or x = −7
Since length cannot be negative,
x = 9
∴ Side of square park = 9 m
3(v). In our village, Proloy babu bought 350 chilli plants for planting in his rectangular land. When he put the plants in rows, he noticed that if he would put 24 plants more than the number of rows in each row, 10 plants would remain excess. Let us write the number of rows.
Solution :
Let number of rows = x
Plants in each row = x + 24
According to the question,
x(x + 24) + 10 = 350
x² + 24x − 340 = 0
x² + 34x − 10x − 340 = 0
x(x + 34) − 10(x + 34) = 0
(x + 34)(x − 10) = 0
x = −34 or x = 10
Since number of rows cannot be negative,
x = 10
∴ Number of rows = 10
3(vi). Joseph takes 5 minutes less time than Kuntal to make a product. Joseph makes 6 products more than Kuntal while working for 6 hours. Let us calculate the number of products Kuntal makes during that time.
Solution :
Let Kuntal takes x minutes to make one product.
Joseph takes (x − 5) minutes
Products made by Kuntal in 6 hours =
360
x
Products made by Joseph in 6 hours =
360
x − 5
According to the question,
360
x − 5
=
360
x
+ 6
360x = 360(x − 5) + 6x(x − 5)
360x = 360x − 1800 + 6x² − 30x
6x² − 30x − 1800 = 0
x² − 5x − 300 = 0
x² − 20x + 15x − 300 = 0
x(x − 20) + 15(x − 20) = 0
(x − 20)(x + 15) = 0
x = 20 or x = −15
Since time cannot be negative,
x = 20
Products made by Kuntal =
360
20
= 18
∴ Kuntal makes 18 products.
3(vii). The speed of a boat in still water is 8 km/hr. If the boat can go 15 km downstream and 22 km upstream in 5 hours, then let us write the speed of the stream.
Solution :
Let speed of stream = x km/hr
Downstream speed = (8 + x) km/hr
Upstream speed = (8 − x) km/hr
According to the question,
15
8 + x
+
22
8 − x
= 5
15(8 − x) + 22(8 + x) = 5(64 − x²)
120 − 15x + 176 + 22x = 320 − 5x²
296 + 7x = 320 − 5x²
5x² + 7x − 24 = 0
5x² + 15x − 8x − 24 = 0
5x(x + 3) − 8(x + 3) = 0
(x + 3)(5x − 8) = 0
x = −3 or
8
5
Since speed cannot be negative,
x =
8
5
∴ Speed of stream =
8
5
km/hr
3(viii). A superfast train runs at the speed of 15 km/hr more than that of an express train. Leaving the same station the superfast train reached a station of 180 km distance one hour before than the express train. Let us determine the speed of the superfast train.
Solution :
Let speed of express train = x km/hr
Speed of superfast train = (x + 15) km/hr
Time taken by express train =
180
x
Time taken by superfast train =
180
x + 15
According to the question,
180
x
−
180
x + 15
= 1
180(x + 15) − 180x = x(x + 15)
2700 = x² + 15x
x² + 15x − 2700 = 0
x² + 60x − 45x − 2700 = 0
x(x + 60) − 45(x + 60) = 0
(x + 60)(x − 45) = 0
x = −60 or x = 45
Since speed cannot be negative,
x = 45
Speed of superfast train = 45 + 15
= 60 km/hr
∴ Speed of superfast train = 60 km/hr
3(ix). Rehana went to the market and saw that the price of dal of 1 kg is ₹20 and the price of rice of 1 kg is ₹40 less than that of price of 1 kg fish. The total quantity of fish and dal each in ₹240 is equal to the quantity of rice in ₹280. Let us calculate the cost price of 1 kg fish.
Solution :
Let price of 1 kg fish = ₹x
Price of 1 kg rice = ₹(x − 40)
Quantity of fish in ₹240 =
240
x
kg
Quantity of dal in ₹240 =
240
20
= 12 kg
Quantity of rice in ₹280 =
280
x − 40
kg
According to the question,
240
x
+ 12 =
280
x − 40
240(x − 40) + 12x(x − 40) = 280x
240x − 9600 + 12x² − 480x = 280x
12x² − 520x − 9600 = 0
3x² − 130x − 2400 = 0
3x² − 180x + 50x − 2400 = 0
3x(x − 60) + 40(x − 60) = 0
(x − 60)(3x + 40) = 0
x = 60 or
x =
−
40
3
Since price cannot be negative,
x = 60
∴ Price of 1 kg fish = ₹60
Let Us Work Out 1.5
1(i). Let us write the nature of the roots of the equation :
2x² + 7x + 3 = 0
Solution :
D = b² − 4ac
= 7² − 4 × 2 × 3
= 49 − 24
= 25
D > 0 and perfect square
∴ Roots are real, unequal and rational.
1(ii). Let us write the nature of the roots of the equation :
3x² − 2√6x + 2 = 0
Solution :
D = (−2√6)² − 4 × 3 × 2
= 24 − 24
= 0
∴ Roots are real and equal.
1(iii). Let us write the nature of the roots of the equation :
2x² − 7x + 9 = 0
Solution :
D = (−7)² − 4 × 2 × 9
= 49 − 72
= −23
D < 0
∴ Roots are imaginary (no real roots).
1(iv). Let us write the nature of the roots of the equation :
2
5
x² −
2
3
x + 1 = 0
Solution :
D =
(
−
2
3
)²
− 4 ×
2
5
× 1
=
4
9
−
8
5
=
20 − 72
45
=
−
52
45
D < 0
∴ Roots are imaginary.
2(i). Find the value of k for which the equation
49x² + kx + 1 = 0
has equal roots.
Solution :
For equal roots,
D = 0
k² − 4 × 49 × 1 = 0
k² − 196 = 0
k² = 196
k = ±14
2(ii). Find the value of k for which the equation
3x² − 5x + 2k = 0
has equal roots.
Solution :
(−5)² − 4 × 3 × 2k = 0
25 − 24k = 0
24k = 25
k =
25
24
2(iii). Find the value of k for which the equation
9x² − 24x + k = 0
has equal roots.
Solution :
(−24)² − 4 × 9 × k = 0
576 − 36k = 0
36k = 576
k = 16
2(iv). Find the value of k for which the equation
2x² + 3x + k = 0
has equal roots.
Solution :
3² − 4 × 2 × k = 0
9 − 8k = 0
k =
9
8
2(v). Find the value of k for which the equation
x² − 2(5 + 2k)x + 3(7 + 10k) = 0
has equal roots.
Solution :
[−2(5 + 2k)]² − 4 × 1 × 3(7 + 10k) = 0
4(5 + 2k)² − 12(7 + 10k) = 0
(5 + 2k)² − 3(7 + 10k) = 0
25 + 20k + 4k² − 21 − 30k = 0
4k² − 10k + 4 = 0
2k² − 5k + 2 = 0
2k² − 4k − k + 2 = 0
2k(k − 2) − 1(k − 2) = 0
(k − 2)(2k − 1) = 0
k = 2 or
k =
1
2
2(vi). Find the value of k for which the equation
(3k + 1)x² + 2(k + 1)x + k = 0
has equal roots.
Solution :
[2(k + 1)]² − 4(3k + 1)k = 0
4(k + 1)² − 4k(3k + 1) = 0
(k + 1)² − k(3k + 1) = 0
k² + 2k + 1 − 3k² − k = 0
2k² − k − 1 = 0
2k² + k − 2k − 1 = 0
k(2k + 1) − 1(2k + 1) = 0
(2k + 1)(k − 1) = 0
k =
−
1
2
or k = 1
3(i). Form the quadratic equation whose roots are 4 and 2.
Solution :
(x − 4)(x − 2) = 0
x² − 6x + 8 = 0
3(ii). Form the quadratic equation whose roots are −4 and −3.
Solution :
(x + 4)(x + 3) = 0
x² + 7x + 12 = 0
3(iii). Form the quadratic equation whose roots are −4 and 3.
Solution :
(x + 4)(x − 3) = 0
x² + x − 12 = 0
3(iv). Form the quadratic equation whose roots are 5 and −3.
Solution :
(x − 5)(x + 3) = 0
x² − 2x − 15 = 0
4. Find the value of m for which the roots of the equation
4x² + 4(3m − 1)x + (m + 7) = 0
are reciprocal to each other.
Solution :
If the roots are reciprocal to each other, then
αβ = 1
For the equation ax² + bx + c = 0,
αβ =
c
a
m + 7
4
= 1
m + 7 = 4
m = −3
∴ Required value of m = −3
5. If two roots of the quadratic equation are equal, then prove that
2b² = a² + c²
for the equation
(b − c)x² + (c − a)x + (a − b) = 0
Solution :
Since the roots are equal,
D = 0
(c − a)² − 4(b − c)(a − b) = 0
c² − 2ac + a² − 4(ab − b² − ac + bc) = 0
c² − 2ac + a² − 4ab + 4b² + 4ac − 4bc = 0
a² + c² + 2ac − 4ab − 4bc + 4b² = 0
a² + c² + 2ac − 4b(a + c) + 4b² = 0
(a + c)² − 4b(a + c) + 4b² = 0
(a + c − 2b)² = 0
a + c = 2b
∴ Proved.
6. If the roots of the equation
(a² + b²)x² − 2(ac + bd)x + (c² + d²) = 0
are equal, then prove that
a
b
=
c
d
Solution :
Since the roots are equal,
D = 0
[−2(ac + bd)]² − 4(a² + b²)(c² + d²) = 0
4(ac + bd)² − 4(a² + b²)(c² + d²) = 0
(ac + bd)² − (a² + b²)(c² + d²) = 0
a²c² + b²d² + 2abcd − a²c² − a²d² − b²c² − b²d² = 0
2abcd − a²d² − b²c² = 0
a²d² − 2abcd + b²c² = 0
(ad − bc)² = 0
ad = bc
a
b
=
c
d
∴ Proved.
7. Let us prove that the quadratic equation
2(a² + b²)x² + 2(a + b)x + 1 = 0
has no real root if a ≠ b.
Solution :
D = [2(a + b)]² − 4 × 2(a² + b²) × 1
= 4(a + b)² − 8(a² + b²)
= 4(a² + 2ab + b²) − 8a² − 8b²
= 4a² + 8ab + 4b² − 8a² − 8b²
= −4a² + 8ab − 4b²
= −4(a² − 2ab + b²)
= −4(a − b)²
Since a ≠ b,
(a − b)² > 0
∴ D < 0
Hence the equation has no real roots.
8. If the roots of the equation
5x² + 2x − 3 = 0
are α and β, determine the following :
Solution :
For the equation 5x² + 2x − 3 = 0,
α + β =
−
2
5
αβ =
−
3
5
8(i). Find α² + β²
Solution :
α² + β² = (α + β)² − 2αβ
=
(
−
2
5
)²
− 2
(
−
3
5
)
=
4
25
+
6
5
=
34
25
8(ii). Find α³ + β³
Solution :
α³ + β³ = (α + β)³ − 3αβ(α + β)
=
(
−
2
5
)³
− 3
(
−
3
5
)
(
−
2
5
)
=
−
8
125
−
18
25
=
−
98
125
8(iii). Find
1
α
+
1
β
Solution :
1
α
+
1
β
=
α + β
αβ
=
−
2
5
−
3
5
=
2
3
8(iv). Find
α²
β
+
β²
α
Solution :
α²
β
+
β²
α
=
α³ + β³
αβ
=
−
98
125
−
3
5
=
98
75
9. If one root of the equation ax² + bx + c = 0 is twice the other, then show that
2b² = 9ac
Solution :
Let roots be α and 2α
α + 2α =
−
b
a
3α =
−
b
a
α =
−
b
3a
Also,
α × 2α =
c
a
2α² =
c
a
2
(
−
b
3a
)²
=
c
a
2b² =
9ac
∴ Proved.
10. Form the equation whose roots are reciprocals of the roots of the equation
x² + px + 1 = 0
Solution :
Let roots of the equation be α and β.
α + β = −p
αβ = 1
Required roots are
1
α
and
1
β
Sum =
α + β
αβ
= −p
Product =
1
αβ
= 1
Required equation :
x² − (sum)x + product = 0
x² + px + 1 = 0
11. Determine the equation whose roots are the squares of the roots of the equation
x² + x + 1 = 0
Solution :
Let roots be α and β.
α + β = −1
αβ = 1
Required roots are α² and β².
α² + β² = (α + β)² − 2αβ
= (−1)² − 2 × 1
= 1 − 2
= −1
α²Î²² = (αβ)²
= 1
Required equation :
x² − (sum)x + product = 0
x² + x + 1 = 0
12. V.S.A.
(A) M.C.Q.
(i) The sum of the two roots of the equation
x² − 6x + 2 = 0
is
(a) 2
(b) −2
(c) 6
(d) −6
Solution :
For the equation ax² + bx + c = 0,
Sum of roots =
−
b
a
=
−
−6
1
= 6
∴ Correct answer = (c) 6
(ii) If the product of the roots of the equation
x² − 3x + k = 10
is −2, then the value of k is
(a) −2
(b) −8
(c) 8
(d) 12
Solution :
x² − 3x + (k − 10) = 0
Product of roots =
k − 10
1
According to the question,
k − 10 = −2
k = 8
∴ Correct answer = (c) 8
(iii) If the two roots of the equation
ax² + bx + c = 0
are real and unequal, then
b² − 4ac
will be
(a) > 0
(b) = 0
(c) < 0
(d) none of these
Solution :
For real and unequal roots,
D = b² − 4ac > 0
∴ Correct answer = (a) > 0
(iv) If the roots of the equation
ax² + bx + c = 0
are equal, then
(a)
−b
2a
(b)
b
2a
(c)
−
b²
4a
(d)
b²
4a
Solution :
For equal roots,
D = 0
b² − 4ac = 0
4ac = b²
c =
b²
4a
∴ Correct answer = (d)
(v) If the roots of the equation
3x² + 8x + 2 = 0
are α and β, then the value of
1
α
+
1
β
is
(a)
−
3
8
(b)
2
3
(c) −4
(d) 4
Solution :
α + β =
−
8
3
αβ =
2
3
1
α
+
1
β
=
α + β
αβ
=
−
8
3
2
3
= −4
∴ Correct answer = (c) −4
(B) Let us identify the following statements as True or False :
(i) The two roots of the equation
x² + x + 1 = 0
are real.
Solution :
D = 1² − 4 × 1 × 1
= 1 − 4
= −3
Since D < 0,
roots are not real.
∴ Statement is False.
(ii) The two roots of the equation
x² − x + 2 = 0
are not real.
Solution :
D = (−1)² − 4 × 1 × 2
= 1 − 8
= −7
Since D < 0,
roots are not real.
∴ Statement is True.
(C) Let us fill in the blanks :
(i) The ratio of the sum and product of the roots of the equation
7x² − 12x + 18 = 0
is ______
Solution :
Sum =
−
−12
7
=
12
7
Product =
18
7
Ratio =
12
7
:
18
7
= 12 : 18
= 2 : 3
∴ Blank = 2 : 3
(ii) If two roots of the equation
ax² + bx + c = 0
are reciprocal to each other, then ______
Solution :
If roots are reciprocal,
αβ = 1
But,
αβ =
c
a
c
a
= 1
a = c
∴ Blank = a = c
(iii) If the roots of the equation
ax² + bx + c = 0
are reciprocal to each other and opposite in sign, then
a + c = ______
Solution :
If roots are opposite in sign,
α + β = 0
Also reciprocal,
αβ = −1
But,
αβ =
c
a
c
a
= −1
a + c = 0
∴ Blank = 0
13(i). Write the quadratic equation whose sum of roots is 14 and product of roots is 24.
Solution :
Required equation :
x² − (sum)x + product = 0
x² − 14x + 24 = 0
13(ii). If the sum and the product of the roots of the equation
kx² + 2x + 3k = 0
are equal, find k.
Solution :
Sum =
−
2
k
Product =
3k
k
= 3
According to the question,
−
2
k
= 3
k =
−
2
3
13(iii). If the roots of the equation
x² − 22x + 105 = 0
are α and β, find the value of (α − β).
Solution :
α + β = 22
αβ = 105
(α − β)² = (α + β)² − 4αβ
= 22² − 4 × 105
= 484 − 420
= 64
α − β = 8
13(iv). If the sum of the roots of the equation
x² − x = k(2x − 1)
is zero, find k.
Solution :
x² − x = 2kx − k
x² − (1 + 2k)x + k = 0
Sum of roots =
1 + 2k
According to the question,
1 + 2k = 0
k =
−
1
2
13(v). If one root of the equations
x² + bx + 12 = 0
and
x² + bx + q = 0
is 2, find q.
Solution :
Since 2 is a root of
x² + bx + 12 = 0
2² + 2b + 12 = 0
4 + 2b + 12 = 0
2b + 16 = 0
b = −8
Now for the equation
x² − 8x + q = 0
Since 2 is a root,
2² − 8(2) + q = 0
4 − 16 + q = 0
q = 12
∴ Required value of q = 12
