Question

If alpha and beta are the zeros of the quadratic polynomial f(x) =kx^2+4x+4 such that Alpha square plus beta square is equals to 24, find the value of k give your answer correctly otherwise the answer will be reported

Answer 1

This is the answer I guess!!!!

Answer 2

Answer:

$k=-1\:Or\:k=\frac{2}{3}$

Step-by-step explanation:

$Given \: \alpha\:and\:\beta\\are \:zeros \:of\:the\\quadratic \: polynomial \\f(x)=kx^{2}+4x+4\:such \:that\\\alpha^{2}+\beta^{2}=24--(1)$

$Compare \: this \: with \\ax^{2}+bx+c ,\:we \:get$

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

$\implies \alpha+\beta=\frac{-4}{k}---(2)$

$Product\:of \:the\: zeroes\\=\frac{c}{a}$

$\implies \alpha\beta=\frac{4}{k}--(3)$

$\alpha^{2}+\beta^{2}=24--(1)$

$\implies \left(\alpha+\beta\right)^{2}-2\alpha\beta=24$

$\implies \left(\frac{-4}{k}\right)^{2}-2\times \frac{4}{k}=24\:[from \:(2) \:and\:(3)]$

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

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

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

/* Divide each term by 8, we get

$\implies 2-k=3k^{2}$

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

/* Splitting the middle term,we get

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

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

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

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

$\implies k=-1\:Or\:3k=2$

$\implies k=-1\:Or\:k=\frac{2}{3}$

Therefore,

$k=-1\:Or\:k=\frac{2}{3}$

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