Question

find the value of a if one root of the equation 8x^2-6x+a=0 is the square of the other..​

Answer 1

Answer:

-4 and 24.

Step-by-step explanation:

Given is the equation 8x^2 - 6x - a - 3 = 0

Its said that one root of this equation is square of the other, so we can assume them as k, k^2.

So,

k + k^2 = -(-6)/8

=> k + k^2 = 3/4  ____________Equation 1

also,

k * k^2 = (- a - 3)/8

=> k^3 = -(a + 3)/8 __________Equation 2

We have to solve these 2 equations and find the values of a. You can easily see that finding k from Equation 1 and then substituting it into Equation 2 will give us the value of a.

k + k^2 = 3/4

=> 4k + 4k^2 = 3

=> 4k^2 + 4k - 3 = 0

=> 4k^2 - 2k + 6k - 3 = 0

=> 2k(2k - 1) + 3(2k - 1) = 0

=> (2k - 1)(2k + 3) = 0

=> (2k - 1) = 0; (2k + 3) = 0

=> k = 1/2; k = -3/2

Putting k = 1/2 in Equation 2, we'll get

(1/2)^3 = -(a + 3)/8

=> 1/8 = -(a + 3)/8

=> 1 = -(a + 3)

=> a = -4

Putting k = -3/2 in Equation 2, we'll get

(-3/2)^3 = -(a + 3)/8

=> -27/8 = -(a + 3)/8

=> 27 = (a + 3)

=> a =24

So, the values of a are -4 and 24.