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High School
Algebra
Algebra I
Quadratic function and equation

FiessyFrimatsd0

Answered question

2022-01-31

Find a such that

{x_{1}}^{2} + {x_{2}}^{2}

takes the minimal value where

x_{1}, x_{2}

are solutions to

x^{2} - a x + (a - 1) = 0

Answer & Explanation

gekraamdbk

Beginner2022-02-01Added 13 answers

(x - x_{1}) (x - x_{2}) = 0

since these are the roots, so:

x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} = 0

x_{1} + x_{2} = a x_{1} x_{2} = a - 1

now we can say:

{(x_{1} + x_{2})}^{2} = x_{1}^{2} + 2 x_{1} x_{2} + x_{2}^{2} = (x_{1}^{2} + x_{2}^{2}) + 2 (a - 1) = a^{2}

which finally gives us:

y = x_{1}^{2} + x_{2}^{2} = a^{2} - 2 a + 2

now to find the minimum differentiate:

\frac{d y}{d a} = 2 a - 2

so for

y^{'} = 0 we get a = 1 giving min (x_{1}^{2} + x_{2}^{2}) = 1

coolbananas03ok

Beginner2022-02-02Added 20 answers

From $x^{2} - a x + (a - 1)$ , we know that any root satisfies
$x^{2} = a x - (a - 1)$
$\begin{aligned} x_{1}^{2} + x_{2}^{2} & = (a x_{1} - (a - 1)) + (a x_{2} - (a - 1)) \\ = a (x_{1} + x_{2}) - 2 a + 2 . \end{aligned}$
From $\begin{aligned} x^{2} - a x + (a - 1) & = (x - x_{1}) (x - x_{2}) \\ = x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2}, \end{aligned}$
we have $x_{1} + x_{2} = a$ , so
$\begin{aligned} x_{1}^{2} + x_{2}^{2} & = a (a) - 2 a + 2 \\ = a^{2} - 2 a + 1 - 1 + 2 \\ = (a - 1)^{2} + 1, \end{aligned}$
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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