## Quadratic Equations – Exercise 9.8 – Questions:

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

iii) -3, ^{3}/_{2}

- iv)
^{2}/_{3},^{3}/_{2} - v) (2 + √3), (2 – √3)
- vi) (-3 + 2√5)(-3 – 2√5)

B.

- If m and n are the roots of the equation x
^{2}– 6x + 2 = 0 find the value of

(i) (m + n)mn

(ii) ^{1}/_{m} + ^{1}/_{n}

(iii) m^{3}n^{2} + n^{3}m^{2}

(iv) ^{1}/_{n} – ^{1}/_{m}

- IF a and b are the roots of the equation 3m
^{2}= 6m + 5, find the value of

(i) ^{a}/_{b} + ^{b}/_{a}

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

- If p and q are the roots of the equation 2a
^{2}– 4a + 1 = 0. Find the value of

(i) (p + q)^{2 }+ 4pq

(ii) p^{3} + q^{3}

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

**Quadratic Equations – Exercise 9.8 – Solution:**

**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 x^{2} –(m+n)x + mn = 0

Thus, x^{2} – 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 x^{2} –(m+n)x + mn = 0

Thus, x^{2} – 1x + (-30) = 0

x^{2} – 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 x^{2} –(m+n)x + mn = 0

Thus, x^{2} – (-^{3}/_{2}) x + (-^{9}/_{2}) = 0

x^{2} + ^{3}/_{2}x – ^{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 x^{2} –(m+n)x + mn = 0

Thus, x^{2} – (^{13}/_{6}) x + (1) = 0

x^{2} – ^{13}/_{6}x + 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 x^{2} –(m+n)x + mn = 0

Thus, x^{2} – (4) x + (1) = 0

x^{2} – 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 x^{2} –(m+n)x + mn = 0

Thus, x^{2} – (-6) x + (-36) = 0

x^{2} + 6x – 36 = 0

**B.**

**If m and n are the roots of the equation x**^{2}– 6x + 2 = 0 find the value of

**(i) (m + n)mn **

**(ii) ^{1}/_{m} + ^{1}/_{n}**

**(iii) m ^{3}n^{2} + n^{3}m^{2}**

**(iv) ^{1}/_{n} – ^{1}/_{m}**

Solution:

The given quadratic equation is x^{2} – 6x + 2 = 0 which is of the form ax^{2} + 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) m^{3}n^{2} + n^{3}m^{2}

= (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

**If a and b are the roots of the equation 3m**^{2}= 6m + 5, find the value of

**(i) ^{a}/_{b} + ^{b}/_{a}**

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

Solution:

The given equation is 3m^{2} = 6m + 5, this can be written as 3m^{2} – 6m – 5 = 0, which is of the form mx^{2} + nx + p = 0 where m = 3 ; n = -6 and p = -5

**If p and q are the roots of the equation 2a**^{2}– 4a + 1 = 0. Find the value of

**(i) (p + q) ^{2 }+ 4pq**

**(ii) p ^{3} + q^{3}**

Solution:

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

**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 x^{2} –(m+n)x + mn = 0

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

pqx^{2} – (p^{2} + q^{2})x – pq = 0

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

Solution:

The given quadratic equation is x^{2} + 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

**Find the value of p so that the equation 2x**^{2}– 3qx + 5q = 0 has one root which is twice the other.

Solution:

The given quadratic equation is 2x^{2} – 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} ⇒ 2m^{2} = (^{5q}/_{2})

⇒ 2(^{q}/_{2})^{2} = ^{5q}/_{2}

⇒ 2(^{q^2}/_{4}) = ^{5q}/_{2}

⇒ ^{q^2}/_{4} = ^{5q}/_{4}

⇒q^{2} = 5q

⇒q = 5

**Find the value of p so that the equation 4x**^{2}– 8px + 9 = 0 has roots whose difference is 4.

Solution:

The given quadratic equation is 4x^{2} – 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}

⇒m^{2} – 4m – ^{9}/_{4} = 0

⇒(p+2)^{2} – 4(p+2) – ^{9}/_{4} = 0

⇒p^{2} + 4p + 4 – 4p – 8 – ^{9}/_{4} = 0

⇒p^{2} –^{25}/_{4} = 0

⇒p^{2} = ^{25}/_{4}

⇒p = ±^{5}/_{2}

**If one root of the equation x**^{2}+ px + q = 0 is 3 times the other prove that 3p^{2}= 16q.

Solution:

The given quadratic equation is x^{2} + 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

⇒3m^{2} = q

⇒3(-^{p}/_{4})^{2} = q

⇒3^{p^2}/_{16} = q

⇒3p^{2} = 16q