Question

if the roots of the equation 8x2-6x-k=0 be the square of the other then prove that K=1 or K= -27

Answer 1

CORRECT QUESTION :–

▪︎ If the roots of the equation 8x²- 6x - k = 0 be the square of the other then prove that K = -1 or K = 27.

ANSWER :–

GIVEN :–

▪︎ A quadratic equation 8x² - 6x - k = 0.

▪︎ Roots of the equation be the square of the other.

TO PROVE :–

Value of K = 1 or K= -27.

SOLUTION :–

▪︎ Let the roots of quadratic equation are a and a².

▪︎ We know that –

• Sum of roots = a + a² = -(b/a)

=> a + a² = 6/8

=> 8a² + 8a - 6 = 0

=> 8a² + 12a - 4a - 6 = 0

=> 4a(2a + 3) - 2(2a + 3) = 0

=> (4a - 2)(2a + 3) = 0

=> a = ½ or a = -(3/2)

• Product of roots = (a)(a²) = (c/a)

=> a³ = -(k/8)

• put a = ½ –

=> (½)³ = -(k/8)

=> k = -1

• put a = -(3/2) –

=> -(3/2)³ = -(k/8)

=> k = 27

Hence proved , k = -1 or k = 27

Answer 2

Given

The roots of the equation are the square of the other

Let the roots of quadratic equation are a and a².

Sum of roots

$\implies a + a^{2} = -(\dfrac{b}{a} )$

$\implies a + a^{2} = \dfrac{6}{8}$

$\implies 8a^{2} + 8a - 6 = 0$

$\implies 8a^{2} + 12a - 4a - 6 = 0$

$\implies 4a(2a + 3) - 2(2a + 3) = 0$

$\implies (4a - 2)(2a + 3) = 0$

Now,

$4a-2=0\\\ \implies a = \dfrac{1}{2}$

$2a+3=0\\\ \implies a = \dfrac{-3}{2}$

Product of roots = c/a

$\implies (\dfrac{1}{2})^{3} = \dfrac{-k}{8}$

$\implies k = 27$

Also

$\implies (\dfrac{-3}{2} )^{3} = \dfrac{-k}{8}$

$\implies k = 1$

Hence proved !!