Let a and b be the roots of kx^2 + 4x + 4 = 0. If a^2 + b^2 = 24, then find all possible values of k.
+9519 By Vieta's formula,
\(a + b = -\dfrac4k\\ ab = \dfrac4k\)
We start with an identity to find the value of a2 + b2.
\(a^2 + 2ab + b^2 = (a + b)^2\\ a^2 + b^2 + 2\left(\dfrac4k\right) = \left(-\dfrac4k\right)^2\\ a^2 + b^2 = \dfrac{16}{k^2} - \dfrac8k =\dfrac{8(2 - k)}{k^2}\)
Then, we equate it to 24.
\(\dfrac{8(2-k)}{k^2} = 24\\ 3k^2 = 2 - k\\ 3k^2 + k - 2 = 0\)
I believe you can do the rest.
