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If $\alpha$ and $\beta(\neq 0)$ are the rootsof the quadratice quation $x^{2}+\alpha x$ $-\beta=0$, then the quadratic expression $-x^{2}+\alpha x+\beta$ where $x \in$ $\begin{array}{ll}\text { R has }\end{array}$
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Verified Answer
The correct answer is:
Greatest value $\frac{9}{4}$
$\alpha+\beta=-\alpha, \alpha \beta=-\beta$
$\Rightarrow \alpha \beta+\beta=0$
$\Rightarrow(\alpha+1) \beta=0$
$\Rightarrow \alpha=-1(\beta \neq 0)$
$\Rightarrow(2 \alpha+\beta)=0$
$\Rightarrow \beta=2$
$\therefore-x^{2}+\alpha x+\beta=-x^{2}-x+2$
Greatest value $=-\frac{1+8}{-4}=\frac{9}{4}$
$\Rightarrow \alpha \beta+\beta=0$
$\Rightarrow(\alpha+1) \beta=0$
$\Rightarrow \alpha=-1(\beta \neq 0)$
$\Rightarrow(2 \alpha+\beta)=0$
$\Rightarrow \beta=2$
$\therefore-x^{2}+\alpha x+\beta=-x^{2}-x+2$
Greatest value $=-\frac{1+8}{-4}=\frac{9}{4}$
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