How do you show that one of the roots of $x^{2} + a x + b = 0$ $x^2+ax+b=0$ is $a$ $a$ , if and only if $b = 0$ $b=0$ ?

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How do you show that one of the roots of $x^{2} + a x + b = 0$ $x^2+ax+b=0$ is $a$ $a$ , if and only if $b = 0$ $b=0$ ?

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Answer 1 · 2018-03-27T23:41:04

First of all, the root would be $x = - a$ $x=-a$ . Treat the equation as if it is a full quadratic, and then attempt to solve when $b \neq 0$ $bne0$ as well as when $b = 0$ $b=0$ to prove it.

Explanation:

Let's write out the quadratic formula first:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

This assumes the equation looks like: $a x^{2} + b x + c$ $ax^2+bx+c$ in this scenario:

$a = 1$ $a=1$
$b = a$ $b=a$
$c = b$ $c=b$

Filling out the equation:

$x = \frac{- a \pm \sqrt{a^{2} - 4 b}}{2}$ $x=(-a+-sqrt(a^2-4b))/(2)$

Now, if $b \neq 0$ $bne0$ , the solution for the zeroes can be given using the above equation. However, if $b = 0$ $b=0$ , we can move forward:

$x = \frac{- a \pm \sqrt{a^{2} - 4 (0)}}{2} = \frac{- a \pm \sqrt{a^{2}}}{2}$ $x=(-a+-sqrt(a^2-4(0)))/(2)=(-a+-sqrt(a^2))/(2)$

$x = \frac{- a \pm a}{2}$ $x=(-a+-a)/2$

$x = {0, - a}$ $x={0,-a}$

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How do you show that one of the roots of $x^{2} + a x + b = 0$ $x^2+ax+b=0$ is $a$ $a$ , if and only if $b = 0$ $b=0$ ?

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Explanation:

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How do you show that one of the roots of $x^{2} + a x + b = 0$ $x^2+ax+b=0$ is $a$ $a$ , if and only if $b = 0$ $b=0$ ?