Question

if α & β are the zeroes of the quadratic polynomial f(x)=kx square +4x+4 such that α square +β square =24 find the value of k

Answer 1

may this solution will help u

Answer 2

Given that, α and β are the zeros of the Quadratic Polynomial f(x) = kx² + 4x + 4. And, (α² + β²) = 24.

As we know that,

Sum and Product of any Quadratic Polynomial are given by :

(α + β) = (–b)/a

(α β) = c/a

So, Sum and Product of given Quadratic Polynomial : kx² + 4x + 4

Sum of zeroes (α + β) = (–4)/k

Product of zeroes (α β) = 4/k

$\rule{100px}{.3ex}$

⭒By using I D E N T I T Y :

$\bigstar\;{\underline{\boxed{\pink{\frak{ \pmb{(a + b)^2 = a^2 + b^2 + 2ab}}}}}}$

Therefore,

$\begin{gathered} \sf :\implies \Big(\alpha + \beta \Big)^2 = \alpha^2 + \beta^2 + 2\alpha \beta \\\\\\:\implies\sf \bigg(\dfrac{-4}{\;k}\bigg)^2 = 24 + 2 \bigg(\dfrac{4}{k}\bigg) \\\\\\:\implies\sf \bigg(\dfrac{4^2}{k^2} \bigg)= 24 + 2 \bigg(\dfrac{4}{k}\bigg) \\\\\\:\implies\sf \bigg(\dfrac{16}{k^2}\bigg) = 24 + 8k \\\\\\:\implies\sf 16 = 24k^2 + 8k \\\\\\:\implies\sf 2 = 3k^2 + k \\\\\\:\implies\sf 3k^2 + k -2 = 0 \\\\\\:\implies\sf 3k^2 + 3k - 2k - 2 = 0 \\\\\\:\implies\sf 3k(k + 1) -2(k + 1) = 0 \\\\\\:\implies\sf (3k - 2) (k + 1) = 0 \\\\\\:\implies\underline{\boxed{\frak{\pmb{\pink{k = \dfrac{2}{3} \:or\; -1}}}}}\;\bigstar\end{gathered}$

$\therefore{\underline{\sf{Hence,\;required\:value\;of\;k\;are\; \sf\pmb{ \dfrac{2}{3},\;-1.}}}}$