Question

If alpha and beta are the zeroes of a quadratic polynomial fx = kx2 + 4x +4 such that a2 + b2 = 24 find the value of k....someone send me the answer fast...

Answer 1

$\red{\tt{\underline{\underline{Answer:}}}}$

${\tt{The \ value \ of \ k \ is \ -1 \ or \ \frac{2}{3}}}$

$\orange{\tt{\underline{\underline{Given:}}}}$

${\tt{The \ given \ polynomial \ is}}$

${\implies{\tt{f(x)=kx^{2}+4x+4}}}$

${\implies{\tt{\alpha \ and \ \beta \ are \ the \ zeroes}}}$

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

$\pink{\tt{\underline{\underline{To \ find:}}}}$

${\tt{The \ value \ of \ k.}}$

$\green{\tt{\underline{\underline{Solution:}}}}$

${\tt{The \ given \ polynomial \ is}}$

${\implies{\tt{f(x)=kx^{2}+4x+4}}}$

${\tt{Here, \ a=k, \ b=4 \ and \ c=4}}$

${\tt{Sum \ of \ zeroes=\frac{-b}{a}}}$

${\tt{\therefore{\alpha+\beta=\frac{-4}{k}...(1)}}}$

${\tt{Product \ of \ zeroes=\frac{c}{a}}}$

${\tt{\therefore{\alpha\beta=\frac{4}{k}..(2)}}}$

${\tt{According \ to \ identity}}$

${\tt{a^{2}+b^{2}=(a+b)^{2}-2ab}}$

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

${\tt{From \ (1) \ and \ (2)}}$

${\tt{(\frac{-4}{k})^{2}-2\times\frac{4}{k}=24}}$

${\tt{\therefore{\frac{16}{k^{2}}-\frac{8}{k}=24}}}$

${\tt{\therefore{\frac{16-8k}{k^{2}}=24}}}$

${\tt{\therefore{16-8k=24k^{2}}}}$

${\tt{\therefore{24k^{2}+8k-16=0}}}$

${\tt{Dividing \ equation \ throughout \ by \ 4}}$

${\tt{\therefore{6k^{2}+2k-4=0}}}$

${\tt{\therefore{6k^{2}+6k-4k-4=0}}}$

${\tt{\therefore{6k(k+1)-4(k+1)=0}}}$

${\tt{\therefore{(k+1)(6k-4)=0}}}$

${\tt{\therefore{k=-1 \ or \ \frac{4}{6}}}}$

${\tt{\therefore{k=-1 \ or \ \frac{2}{3}}}}$

$\purple{\tt{\therefore{The \ value \ of \ k \ is \ -1 \ or \ \frac{2}{3}}}}$

Answer 2

★$\color{darkblue}\underline{\underline{\sf Question}}$

$\rm{If \ \alpha \ and \ \beta \ are \ the \ zeroes \ of \ a \ quadratic}$

$\rm{ polynomial \ f(x) \ = \ k({x}^{2}+4x+4 \ Such}$

$\rm{ that \ {a}^{2}+{b}^{2}=24 \ find \ the \ value \ of \ k}$

_________________________________________

★$\color{purple}\underline{\underline{\sf Answer:}}$

$\rm\pink{The \ value \ of \ k \ is \ -1 \ or \ \frac{2}{3}}$

_________________________________________

$\rm\green{\underline{\underline{Given:}}}$

$\rm{The \ given \ polynomial \ is}$

➜${\rm{f(x)=kx^{2}+4x+4}}$

➜${\rm{\alpha \ and \ \beta \ are \ the \ zeroes}}$

➜${\rm{\alpha^{2}+\beta^{2}=24}}$

_________________________________________

$\orange{\rm{\underline{\underline{To \ find:}}}}$

${\rm{The \ value \ of \ k}}$

_________________________________________

$\green{\rm{\underline{\underline{Solution:}}}}$

${\rm{The \ given \ polynomial \ is}}$

$\rm{f(x)=kx^{2}+4x+4}$

$\rm{Here, \ a=k, \ b=4 \ and \ c=4}$

${\rm{Sum \ of \ zeroes=\frac{-b}{a}}}$

$\rm{\alpha+\beta=\frac{-4}{k}...(a)}$

$\rm{Product \ of \ zeroes \ = \ \frac{c}{a}}$

$\rm{\alpha\beta=\frac{4}{k}...(b)}$

${\rm{a^{2}+b^{2}=(a+b)^{2}-2ab}}$

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

${\rm{From \ (a) \ and \ (b)}}$

${\rm{(\frac{-4}{k})^{2}-2\times\frac{4}{k}=24}}$

${\rm{\frac{16}{k^{2}}-\frac{8}{k}=24}}$

${\rm{\frac{16-8k}{k^{2}}=24}}$

${\rm{16-8k=24k^{2}}}$

⟾ ${\rm{24k^{2}+8k-16=0}}$

⟾ ${\rm{6k^{2}+2k-4=0}}$

⟾ ${\rm{6k^{2}+6k-4k-4=0}}$

⟾ ${\rm{6k(k+1)-4(k+1)=0}}$

⟾ ${\rm{(k+1)(6k-4)=0}}$

⟾${\rm{k=-1 \ or \ \frac{4}{6}}}$

⟾${\rm{k=-1 \ or \ \frac{2}{3}}}$

$\orange{\rm{The \ value \ of \ k \ is \ -1 \ or \ \frac{2}{3}}}$

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$\rm\pink{Hope \ it \ helps \ :))}$