FiessyFrimatsd0

2022-01-31

Find a such that ${\left\{{x}_{1}\right\}}^{2}+{\left\{{x}_{2}\right\}}^{2}$ takes the minimal value where ${x}_{1},{x}_{2}$ are solutions to ${x}^{2}-ax+\left(a-1\right)=0$.

gekraamdbk

Expert

$\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)=0$
since these are the roots, so:
${x}^{2}-\left({x}_{1}+{x}_{2}\right)x+{x}_{1}{x}_{2}=0$
so
now we can say: ${\left({x}_{1}+{x}_{2}\right)}^{2}={x}_{1}^{2}+2{x}_{1}{x}_{2}+{x}_{2}^{2}=\left({x}_{1}^{2}+{x}_{2}^{2}\right)+2\left(a-1\right)={a}^{2}$
which finally gives us:
$y={x}_{1}^{2}+{x}_{2}^{2}={a}^{2}-2a+2$
now to find the minimum differentiate:
$\frac{dy}{da}=2a-2$
so for .

coolbananas03ok

Expert

From ${x}^{2}-ax+\left(a-1\right)$, we know that any root satisfies
${x}^{2}=ax-\left(a-1\right)$
$\begin{array}{rl}{x}_{1}^{2}+{x}_{2}^{2}& =\left(a{x}_{1}-\left(a-1\right)\right)+\left(a{x}_{2}-\left(a-1\right)\right)\\ & =a\left({x}_{1}+{x}_{2}\right)-2a+2\text{.}\end{array}$
From $\begin{array}{rl}{x}^{2}-ax+\left(a-1\right)& =\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)\\ & ={x}^{2}-\left({x}_{1}+{x}_{2}\right)x+{x}_{1}{x}_{2}\text{,}\end{array}$
we have ${x}_{1}+{x}_{2}=a$, so
$\begin{array}{rl}{x}_{1}^{2}+{x}_{2}^{2}& =a\left(a\right)-2a+2\\ & ={a}^{2}-2a+1-1+2\\ & =\left(a-1{\right)}^{2}+1\text{,}\end{array}$
which is a nonnegative term plus 1, so is minimized when the nonnegative term is zero, that is, when $a=1$.

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