If α and β are the real roots of the equation x2−(k−2)x+(k2+3k+5)=0 (k∈R) then maximum value of α2+β2 is

α+β=k−2, αβ=k2+3k+5α2+β2=(α+β)2−2αβ=(k−2)2−2k2−6k−10 = −k2−10k−6 Let f(k)=−k2−10k−6 f1(k)=0 ⇒ −2k−10=0 ⇒ k=−5 f11(k)=−2<0Maximum value f(−5)=−25+50−6 = 19

If α and β are the real roots of the equation x2−(k−2)x+(k2+3k+5)=0 (k∈R) then maximum value of α2+β2 is

α+β=k−2, αβ=k2+3k+5α2+β2=(α+β)2−2αβ=(k−2)2−2k2−6k−10 = −k2−10k−6 Let f(k)=−k2−10k−6 f1(k)=0 ⇒ −2k−10=0 ⇒ k=−5 f11(k)=−2<0Maximum value f(−5)=−25+50−6 = 19

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If $α$ and $β$ are the real roots of the equation $x^{2} - (k - 2) x + (k^{2} + 3 k + 5) = 0 (k \in R)$ then maximum value of $α^{2} + β^{2}$ is

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Detailed Solution

$\begin{array}{l} α + β = k - 2, α β = k^{2} + 3 k + 5 \\ α^{2} + β^{2} = {(α + β)}^{2} - 2 α β = {(k - 2)}^{2} - 2 k^{2} - 6 k - 10 \\ = - k^{2} - 10 k - 6 \\ L e t f (k) = - k^{2} - 10 k - 6 \\ f^{1} (k) = 0 \Rightarrow - 2 k - 10 = 0 \Rightarrow k = - 5 \\ f^{11} (k) = - 2 < 0 \end{array}$