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 :

[tex]{ \boldsymbol{k}} \: = \: (- 1 ), \: \bigg(\dfrac{2}{3} \bigg) [/tex]

Explanation :

Given polynomial,

[tex]p(x) = kx^{2} + 4x + 4[/tex]

Where,

[tex] { \alpha }^{2} + { \beta }^{2} = 24[/tex]

Find the value of [tex]\boldsymbol{k}[/tex]

On comparing the polynomial with [tex]ax^2 + bx + c[/tex]

We have,

• a = k
• b = 4
• c = 4

Here,

[tex] \bull \: \alpha + \beta = \dfrac{ – b}{a} = \dfrac{ – 4}{k} [/tex]

[tex] \bull \: \alpha \beta = \dfrac{c}{a} = \dfrac{4}{k} [/tex]

Using identity,

[tex] {( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 { \alpha \beta }[/tex]

On substituting the values,

[tex] \implies \: \bigg( \dfrac{ – 4}{k} \bigg)^{2} = 24 + 2 \bigg( \dfrac{4}{k} \bigg) \\[/tex]

[tex]\implies \: \dfrac{16}{ {k}^{2} } = 24 + \dfrac{8}{k} \\[/tex]

[tex]\implies \: \: 16 = 24 {k}^{2} + 8k \\[/tex]

[tex]\implies \: 0 = {24k}^{2} + 8k – 16 \\[/tex]

[tex]\implies \: 0 = {3k}^{2} + k – 2 \\[/tex]

[tex]\implies \: 0 = {3k}^{2} + 3 k – 2k – 2 \\[/tex]

[tex]\implies \: 0 = 3k(k + 1) – 2(k + 1) \\[/tex]

[tex]\implies \: 0 = (k + 1)(3k – 2) \\[/tex]

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