Class 10 SELINA Solutions Maths Chapter 5 - Quadratic Equations
Quadratic Equations Exercise TEST YOURSELF
Solution 1(a)
Correct Option: (ii) –4
Consider (3x – 5)(x + 3)
= 3x2 – 5x + 9x – 15
= 3x2 + 4x – 15
Now, 3x2 – kx – 15 = (3x – 5)(x + 3)
That is 3x2 – kx – 15 = 3x2 + 4x – 15
Therefore, –k = 4 ⇒ k = –4
Solution 1(b)
Correct Option: (ii) 4
For a quadratic equation ax2 + bx + c = 0, the roots are real and equal if b2 – 4ac = 0.
For kx2 + kx + 1 = 0,
a = k, b = k and c = 1
Now, b2 – 4ac = 0
k2 – 4(k)(1) = 0
k2 – 4k = 0
k(k – 4) = 0
k = 0 or k – 4 = 0 ⇒ k = 4
But for k = 0, the given equation is not satisfied.
Hence, k = 4
(k = 4 is obtained by considering the roots of the equation as real and equal and NOT real and distinct)
Solution 1(c)
Correct Option: (iii) 5 or –1
x2 – 4x = 5
x2 – 4x – 5 = 0
x2 – 5x + x – 5 = 0
x(x – 5) + 1(x – 5) = 0
(x – 5)(x + 1) = 0
x = 5 or x = –1
Solution 1(d)
Correct Option: (iv) 0 or 7
x2 – 7x = 0
x(x – 7) = 0
x = 0 or x – 7 = 0
x = 0 or x = 7
Solution 1(e)
Correct Option: (i) 1
Substituting x = 1 in ,
Solution 2
Given i.e
So, the given quadratic equation becomes
Hence, the values of x are and.
Solution 3
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
Solution 4
Solution 5
Given quadratic equation is
(i) When the equation has no roots
(ii) When the roots of are
or
Solution 6
Given quadratic equation is
Using quadratic formula,
⇒ x = a + 1 or x = -a - 2 = -(a + 2)
Solution 7
Given quadratic equation is
Since, m and n are roots of the equation, we have
and
Hence, .
Solution 8
Given quadratic equation is …. (i)
One of the roots of (i) is , so it satisfies (i)
So, the equation (i) becomes
Hence, the other root is.
Solution 9
Given quadratic equation is …. (i)
One of the roots of (i) is -3, so it satisfies (i)
Hence, the other root is 2a.
Solution 10
Given quadratic equation is ….. (i)
Also, given and
and
So, the equation (i) becomes
Hence, the solution of given quadratic equation are and.
Solution 11
Given quadratic equation is …. (i)
The quadratic equation has equal roots if its discriminant is zero
When , equation (i) becomes
When , equation (i) becomes
∴ x =
Solution 12
Solution 13
Solution 14
Consider the given equation:
Solution 15
Given quadratic equation is …. (i)
The quadratic equation has real roots if its discriminant is greater than or equal to zero
Hence, the given quadratic equation has real roots for.
Solution 16
(i) Given quadratic equation is
D = b2 - 4ac = = 25 - 24 = 1
Since D > 0, the roots of the given quadratic equation are real and distinct.
Using quadratic formula, we have
or
(ii) Given quadratic equation is
D = b2 - 4ac = = 16 - 20 = - 4
Since D < 0, the roots of the given quadratic equation does not exist.
Solution 17
Since, -2 is a root of the equation 3x2 + 7x + p = 1.
⇒ 3(-2)2 + 7(-2) + p = 1
⇒ 12 - 14 + p = 1
⇒ p = 3
The quadratic equation is x2 + k(4x + k - 1) + p = 0
i.e. x2 + 4kx + k2 - k + 3 = 0
Comparing equation x2 + 4kx + k2 - k + 3 = 0 with ax2 + bx + c = 0, we get
a = 1, b = 4k and c = k2 - k + 3
Since, the roots are equal.
⇒ b2 - 4ac = 0
⇒ (4k)2 - 4(k2 - k + 3) = 0
⇒ 16k2 - 4k2 + 4k - 12 = 0
⇒ 12k2 + 4k - 12 = 0
⇒ 3k2 + k - 3 = 0
By quadratic formula, we have
Quadratic Equations Exercise Ex. 5(A)
Solution 2(i)
5x2 - 8x = -3(7 - 2x)
⇒ 5x2 - 8x = 6x - 21
⇒ 5x2 - 14x + 21 =0; which is of the form ax2 + bx + c = 0.
∴ Given equation is a quadratic equation.
Solution 2(ii)
(x - 4)(3x + 1) = (3x - 1)(x +2)
⇒ 3x2 + x - 12x - 4 = 3x2 + 6x - x - 2
⇒ 16x + 2 =0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
Solution 2(iii)
7x3 - 2x2 + 10 = (2x - 5)2
⇒ 7x3 - 2x2 + 10 = 4x2 - 20x + 25
⇒ 7x3 - 6x2 + 20x - 15 = 0; which is not of the form ax2 + bx + c = 0.
∴ Given equation is not a quadratic equation.
Solution 3(i)
x2 - 2x - 15 = 0
For x = 5 to be solution of the given quadratic equation it should satisfy the equation.
So, substituting x = 5 in the given equation, we get
L.H.S = (5)2 - 2(5) - 15
= 25 - 10 - 15
= 0
= R.H.S
Hence, x = 5 is a solution of the quadratic equation x2 - 2x - 15 = 0.
Solution 3(ii)
2x2 - 7x + 9 = 0
For x = -3 to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = 5 in the given equation, we get
L.H.S=2(-3)2 - 7(-3) + 9
= 18 + 21 + 9
= 48
≠ R.H.S
Hence, x = -3 is not a solution of the quadratic equation 2x2 - 7x + 9 = 0.
Solution 4
For x = to be solution of the given quadratic equation it should satisfy the equation
So, substituting x = in the given equation, we get
Solution 5
For x = and x = 1 to be solutions of the given quadratic equation it should satisfy the equation
So, substituting x = and x = 1 in the given equation, we get
Solving equations (1) and (2) simultaneously,
Solution 1(b)
Correct Option: (iii) 3 or –5
(x – 3)(x + 5) = 0
(x – 3) = 0 or (x + 5) = 0
x = 3 or x = –5
Solution 1(c)
Correct Option: (ii) –3
Substituting x = 4 in equation x2 + kx – 4 = 0,
(4)2 + k(4) – 4 = 0
16 + 4k – 4 = 0
4k + 12 = 0
4k = –12
k = –3
Solution 1(d)
Correct Option: (i) –2
Substituting x = 2 in equation 2x2 – 3x + k = 0,
2(2)2 – 3(2) + k = 0
8 – 6 + k = 0
2 + k = 0
k = –2
Solution 1(e)
Correct Option: (iv) 0 or 7
x2 – 7x = 0
x(x – 7) = 0
x = 0 or x – 7 = 0
x = 0 or x = 7
Solution 1(a)
Correct Option: (iv)
4x2 – 9 = 0
4x2 = 9
Quadratic Equations Exercise Ex. 5(B)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16(i)
Solution 16(ii)
Solution 17
If a + 7 =0, then a = -7
and b + 10 =0, then b = - 10
Put these values of a and b in the given equation
Solution 18
4(2x+3)2 - (2x+3) - 14 =0
Put 2x+3 = y
Solution 19
or x = -(a + b)
Solution 20
Solution 1(a)
Correct Option: (iii) –1 or 7
x2 – 6x – 7 = 0
x2 – 7x + x – 7 = 0
x(x – 7) + (x – 7) = 0
(x – 7)(x + 1) = 0
x = 7 or x = –1
Solution 1(b)
Correct Option: (ii) –6 or –2
x(x + 8) + 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
Solution 1(c)
Correct Option: (ii) 2
Substituting x = 2 in equation (p – 3)x2 + x + p = 0 is 2,
(p – 3)22 + 2 + p = 0
4p – 12 + 2 + p = 0
5p – 10 = 0
p = 2
Solution 1(d)
Correct Option: (iii) 2 or ½
x2 + 1 = 2.5x
x2 – 2.5x + 1 = 0
x2 – 2x – 0.5x + 1 = 0
x(x – 2) – 0.5(x – 2) = 0
(x – 2)(x – 0.5) = 0
x = 2 or x = 0.5 = ½
Solution 1(e)
Correct Option: (i) x ≠ 0
If x = 0, the term in the equation will be undefined.
Hence, for a quadratic equation , x ≠ 0.
Quadratic Equations Exercise Ex. 5(C)
Solution 2
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
4x2 - 5x - 3 = 0
Here, a = 4, b = -5 and c = -3
Solution 4
Solution 5
Solution 6
Consider the given equation:
Solution 7
Solution 1(a)
Correct Option: (ii) 2.0 or 1.0
x2 – 3x + 2 = 0
Then, a = 1, b = –3, c = 2
Solution 1(b)
Correct Option: (i) 5.00 or –1.00
x2 – 4x – 5 = 0
Then, a = 1, b = –4, c = –5
Solution 1(c)
Correct Option: (ii) 9 or –1
x2 – 8x – 9 = 0
Then, a = 1, b = –8, c = –9
Solution 1(d)
Correct Option: (iii) 3.0 or –1.0
x2 – 2x – 3 = 0
Then, a = 1, b = –2, c = –3
Solution 1(e)
Correct Option: (i) 8 or –2
x2 – 6x – 16 = 0
Then, a = 1, b = –6, c = –16
Quadratic Equations Exercise Ex. 5(D)
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 1(a)
Correct Option: (i) distinct and real roots
2x2 – 3x + 1 = 0
Here, a = 2, b = –3, c = 1
b2 – 4ac = (–3)2 – 4(2)(1) = 9 – 8 = 1 > 0
Hence, the roots are distinct and real.
Solution 1(b)
Correct Option: (i) x2 – 5x + 6 = 0
For a quadratic equation ax2 + bx + c = 0, the roots are real and distinct if b2 – 4ac > 0.
For, x2 – 5x + 6 = 0
a = 1, b = –5, c = 6
b2 – 4ac = (–5)2 – 4(1)(6) = 25 – 24 = 1 > 0
Hence, the roots are distinct and real.
Solution 1(c)
Correct Option: (iv) 4 or –4
For, x2 – px + 4 = 0, the roots are equal.
Then, b2 – 4ac = 0
Here, a = 1, b = –p, c = 4
b2 – 4ac = 0
(–p)2 – 4(1)(4) = 0
p2 – 16 = 0
p2 = 16
p = ±4
Solution 1(d)
Correct Option: (ii) 2x2 – 5x + 9 = 0
For a quadratic equation ax2 + bx + c = 0, the roots are imaginary if b2 – 4ac < 0.
For, 2x2 – 5x + 9 = 0
a = 2, b = –5, c = 9
b2 – 4ac = (–5)2 – 4(2)(9) = –47 < 0
Hence, the roots are imaginary.
Solution 1(e)
Correct Option: (i) 7
Substituting x = 1 in equation 3x2 – mx + 4 = 0,
3(1)2 – m(1) + 4 = 0
3 – m + 4 = 0
m = 7
Quadratic Equations Exercise Ex. 5(E)
Solution 1(a)
Correct Option: (ii) ±1 or ±2
x4 – 5x2 + 4 = 0
Let x2 = y
Then,
y2 – 5y + 4 = 0
y2 – 4y – y + 4 = 0
y(y – 4) – 1(y – 4) = 0
(y – 4)(y – 1) = 0
y = 4 or y = 1
x2 = 4 or x2 = 1
x = ±2 or x = ±1
Solution 1(b)
Correct Option: (ii) 10
20x – 50 = 3x2 – 15x
3x2 – 35x + 50 = 0
3x2 – 30x – 5x + 50 = 0
3x(x – 10) – 5(x – 10) = 0
(x – 10)(3x – 5) = 0
x = 10 or
Solution 1(c)
Correct Option: (iii) x ≠ 3 and x ≠ –5
Here, x – 3 ≠ 0 and x + 5 ≠ 0
x ≠ 3 and x ≠ –5
Solution 1(d)
Correct Option: (i) –2 and 9 or –9 and 2
Since the product is negative, one integer should be negative.
From the given options, option (i) satisfies the condition.
Hence, the two integers are –2 and 9 or –9 and 2.
Solution 2
Solution 3
Solution 4 (i)
Solution 4 (ii)
(x2 - 3x)2 - 16(x2 - 3x) - 36 = 0
Let x2 - 3x = y
Then y2 - 16y - 36 = 0
⇒ y2 - 18y + 2y - 36 = 0
⇒ y(y - 18) + 2(y - 18) = 0
⇒ (y - 18) (y + 2) = 0
If y - 18 = 0 or y + 2 = 0
⇒ x2 - 3x - 18 = 0 or x2 - 3x + 2 = 0
⇒ x2 - 6x + 3x - 18 = 0 or x2 - 2x - x + 2 = 0
⇒ x(x - 6) +3(x - 6) = 0 or x(x - 2) -1(x - 2) = 0
⇒ (x - 6) (x + 3) = 0 or (x - 2) (x - 1) = 0
If x - 6 = 0 or x + 3 = 0 or x - 2 = 0 or x - 1 = 0
then x = 6 or x = -3 or x = 2 or x = 1
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
∴ Given equation reduces to
⇒ 2y2 - 3 = 5y
⇒ 2y2 - 5y - 3 = 0
⇒ 2y2 - 6y + y - 3 = 0
⇒ 2y(y - 3) + 1(y - 3) = 0
⇒ (y - 3)(2y + 1) = 0
⇒ y = 3 and
When, y = 3
⇒ 2x - 1 = 3x + 9
⇒ x = -10
When,
⇒ 4x - 2 = -x - 3