Question

if alfa and bita are zeros of polynomial p(x)=kx^2+4x+4
,such alfa^2+bita^2=24
,find value of k​

Answer 1

Answer :

${ \boldsymbol{k}} \: = \: (- 1 ), \: \bigg(\dfrac{2}{3} \bigg)$

Explanation :

Given polynomial,

$p(x) = kx^{2} + 4x + 4$

Where,

${ \alpha }^{2} + { \beta }^{2} = 24$

Find the value of $\boldsymbol{k}$

On comparing the polynomial with $ax^2 + bx + c$

We have,

• a = k
• b = 4
• c = 4

Here,

$\bull \: \alpha + \beta = \dfrac{ - b}{a} = \dfrac{ - 4}{k}$

$\bull \: \alpha \beta = \dfrac{c}{a} = \dfrac{4}{k}$

Using identity,

${( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 { \alpha \beta }$

On substituting the values,

$\implies \: \bigg( \dfrac{ - 4}{k} \bigg)^{2} = 24 + 2 \bigg( \dfrac{4}{k} \bigg)$

$\implies \: \dfrac{16}{ {k}^{2} } = 24 + \dfrac{8}{k}$

$\implies \: \: 16 = 24 {k}^{2} + 8k$

$\implies \: 0 = {24k}^{2} + 8k - 16$

$\implies \: 0 = {3k}^{2} + k - 2$

$\implies \: 0 = {3k}^{2} + 3 k - 2k - 2$

$\implies \: 0 = 3k(k + 1) - 2(k + 1)$

$\implies \: 0 = (k + 1)(3k - 2)$

${ \underline{\therefore {\sf{The \: value \: of \: { \boldsymbol{k}} \: is \: (- 1 ), \: \bigg(\dfrac{ 2}{3} \bigg) }}}}$