Quadratic Equation Exercise 9.8 – Class X

Quadratic Equations – Exercise 9.8 – Questions:

1. Form the quadratic equation whose roots are
2. i) 3, 5
3. ii) 6, -5

iii) -3, ³/₂

1. iv) ²/₃ , ³/₂
2. v) (2 + √3), (2 – √3)
3. vi) (-3 + 2√5)(-3 – 2√5)

B.

1. If m and n are the roots of the equation x² – 6x + 2 = 0 find the value of

(i) (m + n)mn

(ii) ¹/_m + ¹/_n

(iii) m³n² + n³m²

(iv) ¹/_n – ¹/_m

2. IF a and b are the roots of the equation 3m² = 6m + 5, find the value of

(i) ^a/_b + ^b/_a

(ii) (a + 2b)(2a + b)

3. If p and q are the roots of the equation 2a² – 4a + 1 = 0. Find the value of

(i) (p + q)² + 4pq

(ii) p³ + q³

4. Form a quadratic equation whose roots are ^p/_q and ^q/_p
5. Find the value of ‘k’ so that the equation x² + 4x + (k +2) = 0 has one root which is twice the other.
6. Find the value of p so that the equation 2x² – 3qx + 5q = 0 has one root which is twice the other.
7. Find the value of p so that the equation 4x² – 8px + 9 = 0 has roots whose difference is 4.
8. If one root of the equation x² + px + q = 0 is 3 times the other prove that 3p² = 16q.

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Quadratic Equations – Exercise 9.8 – Solution:

1. Form the quadratic equation whose roots are

i) 3, 5

Solution:

Let m and n be the roots

Then m = 3 and n = 5

Sum of the roots = m + n = 3 + 5 = 8

Product of the roots = mn = 3 x 5 = 15

Standard form is x² –(m+n)x + mn = 0

Thus, x² – 8x + 15 = 0

 

ii) 6, -5

Solution:

Let m and n be the roots

Then m = 6 and n = -5

Sum of the roots = m + n = 6 – 5 = 1

Product of the roots = mn = 6 x (-5) = -30

Standard form is x² –(m+n)x + mn = 0

Thus, x² – 1x + (-30) = 0

x² – x – 30 = 0

 

iii) -3, ³/₂

Solution:

Let m and n be the roots

Then m = -3 and n = ³/₂

Sum of the roots = m + n = -3 + ³/₂ = ^-6+3/₂ = –³/₂

Product of the roots = mn = -3 x ³/₂ = ^-9/₂

Standard form is x² –(m+n)x + mn = 0

Thus, x² – (-³/₂) x + (-⁹/₂) = 0

x² + ³/₂x – ⁹/₂ = 0

 

iv) ²/₃ , ³/₂

Solution:

Let m and n be the roots

Then m = ²/₃ and n = ³/₂

Sum of the roots = m + n = ²/₃ + ³/₂ = ⁴⁺⁹/₆ = ¹³/₆

Product of the roots = mn = ²/₃ x ³/₂ =1

Standard form is x² –(m+n)x + mn = 0

Thus, x² – (¹³/₆) x + (1) = 0

x² – ¹³/₆x + 1 = 0

 

v) (2 + √3), (2 – √3)

Solution:

Let m and n be the roots

Then m = 2 + √3 and n = 2 – √3

Sum of the roots = m + n = (2 + √3)+( 2 – √3) = 4

Product of the roots = mn = (2 + √3)( 2 – √3) = 4 – 2√3 + 2√3 – 3 = 4 – 3 = 1

Standard form is x² –(m+n)x + mn = 0

Thus, x² – (4) x + (1) = 0

x² – 4x + 1 = 0

 

vi) (-3 + 2√5)(-3 – 2√5)

Solution:

Let m and n be the roots

Then m = -3 + 2√5 and n = -3 – 2√5

Sum of the roots = m + n = (-3 + 2√5)+( -3 – 2√5) = -6

Product of the roots = mn = (-3 + 2√5)( -3 – 2√5) = 9 + 6√5 – 6√5 – 4×5 = 9 – 45 = – 36

Standard form is x² –(m+n)x + mn = 0

Thus, x² – (-6) x + (-36) = 0

x² + 6x – 36 = 0

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B.

1. If m and n are the roots of the equation x² – 6x + 2 = 0 find the value of

(i) (m + n)mn

(ii) ¹/_m + ¹/_n

(iii) m³n² + n³m²

(iv) ¹/_n – ¹/_m

Solution:

The given quadratic equation is x² – 6x + 2 = 0 which is of the form ax² + bx + c = 0 where a = 1, b = -6 and c = 2

Therefore, m = 3 + √7 and n = 3 – √7

(i) (m + n)mn

= (3+ √7+ 3 – √7)( 3 + √7)( 3 – √7)

= 6 (3 + √7)( 3 – √7)

=6 (9 – 3√7 + 3√7 – 7)

= 6 x 2

= 12

 

(ii) ¹/_m + ¹/_n

= ¹/_{(3 + √7)} + ¹/_{(3 – √7)}

= ^{(3 – √7)+(3 + √7)}/_{(3 – √7)(3+√7)}

= ⁶/₂

= 3

 

(iii) m³n² + n³m²

= (3 + √7)³( 3 – √7)² + (3 – √7)³( 3 + √7)²

= (90 + 34√7)(16 – 6√7) + (90 – 34√7)(16 + 6√7)

= 12 + 4√7 + 12 – 4√7

= 24

 

(iv) ¹/_n – ¹/_m

= ¹/_{(3 – √7)} – ¹/_{(3 + √7)}

= ^{(3 + √7)-(3 – √7)}/_{(3 – √7)(3+√7)}

= ^2√7/₂

= √7

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2. If a and b are the roots of the equation 3m² = 6m + 5, find the value of

(i) ^a/_b + ^b/_a

(ii) (a + 2b)(2a + b)

Solution:

The given equation is 3m² = 6m + 5, this can be written as 3m² – 6m – 5 = 0, which is of the form mx² + nx + p = 0 where m = 3 ; n = -6 and  p = -5

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3. If p and q are the roots of the equation 2a² – 4a + 1 = 0. Find the value of

(i) (p + q)² + 4pq

(ii) p³ + q³

Solution:

Given p and q are the roots of the equation is 2a² – 4a + 1 = 0, which is of the form ax² + bx + c = 0 where a = 2, b = -4 , x = 1

4. Form a quadratic equation whose roots are ^p/_q and ^q/_p

Solution:

Let ^p/_q and ^q/_p be the roots of a quadratic equation

Sum of the roots = m + n = ^p/_q + ^q/_p = ^{p^2+q^2}/_pq

Product of the roots = mn = ^p/_q x ^q/_p = 1

Standard form is x² –(m+n)x + mn = 0

Thus, x² – (^{p^2+q^2}/_pq) x + (1) = 0

pqx² – (p² + q²)x – pq = 0

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5. Find the value of ‘k’ so that the equation x² + 4x + (k +2) = 0 has one root equal to zero.

Solution:

The given quadratic equation is x² + 4x + (k + 2) = 0 where a = 1, b = 4 , c = k+2

The sum of the roots m+n = –^b/_a = – ⁴/₁ = -4

The product of the roots mn = ^c/_a = ^k+2/₁ = k + 2

If one root is m then other root is zero.

Thus, m = m and n = 0

Therefore, m + n = – 4 ⇒ m + 0 = -4 ⇒m = -4

mn = k + 2 ⇒ 0 = k + 2 ⇒ k = -2

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6. Find the value of p so that the equation 2x² – 3qx + 5q = 0 has one root which is twice the other.

Solution:

The given quadratic equation is 2x²  – 3qx + 5q = 0 where a = 2, b = -3q , c = 5q

The sum of the roots m+n = –^(-3q)/₂ = ^3q/₂

The product of the roots mn = ^c/_a = ^5q/₂

If one root is m then other root is 2m.

Thus, m = m and n = 2m

Therefore, m + 2m = ^3q/₂ ⇒ 3m = ^3q/₂ ⇒m = ^q/₂

mn = ^5q/₂ ⇒ 2m² = (^5q/₂)

⇒ 2(^q/₂)² = ^5q/₂

⇒ 2(^{q^2}/₄) = ^5q/₂

⇒ ^{q^2}/₄ = ^5q/₄

⇒q² = 5q

⇒q = 5

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7. Find the value of p so that the equation 4x² – 8px + 9 = 0 has roots whose difference is 4.

Solution:

The given quadratic equation is 4x² – 8px + 9 = 0 where a = 4, b = -8p , c = 9

The sum of the roots m+n = –^(-8p)/₄ = 2p

The product of the roots mn = ^c/_a = ⁹/₄

If one root is m then other root is (m – 4).

Thus, m = m and n = (m – 4)

m+n = 2p⇒ m+m-4 = 2p⇒ 2m = 2p + 4⇒m = p + 2

mn = ⁹/₄ ⇒m(m – 4) = ⁹/₄

⇒m² – 4m – ⁹/₄ = 0

⇒(p+2)² – 4(p+2) – ⁹/₄ = 0

⇒p² + 4p + 4 – 4p – 8 – ⁹/₄ = 0

⇒p² –²⁵/₄ = 0

⇒p² = ²⁵/₄

⇒p = ±⁵/₂

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8. If one root of the equation x² + px + q = 0 is 3 times the other prove that 3p² = 16q.

Solution:

The given quadratic equation is x² + px + q = 0 where a = 1, b = p , c = q

The sum of the roots m+n = –^(p)/₁ = -p

The product of the roots mn = ^q/₁ = q

If one root is m then other root is 3m.

Thus, m = m and n = 3m

m + n = -p ⇒ m+3m = -p

⇒4m = -p

⇒ m = ^-p/₄

mn = q ⇒m(3m) = q

⇒3m² = q

⇒3(-^p/₄)² = q

⇒3^{p^2}/₁₆ = q

⇒3p² = 16q

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