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How does one factor $9 x - 8 = x^{2}$ $9x - 8 = x^2$ ?

3 Answers

Jul 15, 2016

$(x - 8) (x - 1) = 0$ $(x-8)(x-1)=0$

Explanation:

Given $9 x - 8 = x^{2}$ $9x-8=x^2$

Re-arranging the terms:
$XXX x^{2} - 9 x + 8 = 0$ $color(white)("XXX")x^2-9x+8=0$

with factors:
$XXX (x - 8) (x - 1) = 0$ $color(white)("XXX")(x-8)(x-1)=0$

Jul 15, 2016

$x^{2} - 9 x + 8 = (x - 1) (x - 8)$ $x^2-9x+8=(x-1)(x-8)$

Explanation:

$9 x - 8 = x^{2}$ $9x-8=x^2$

$x^{2} - 9 x + 8 = 0$ $x^2-9x+8=0$

$x^{2} - 9 x + 8 = (x - 1) (x - 8)$ $x^2-9x+8=(x-1)(x-8)$

Jul 15, 2016

$(x - 1) (x - 8) = 0$ $(x-1)(x-8) = 0$

The equation can be solved to give $x = 1, x = 8$ $x = 1, x = 8$

Explanation:

As it is a quadratic trinomial, make it equal to 0, with $+ x^{2}$ $+x^2$

$x^{2} - 9 x + 8 = 0$ $x^2-color(orange)9x color(magenta)(+) color(blue)8 = 0$

Find factors of $8$ $color(blue)8$ which $a d d$ $color(magenta)(add)$ up to $9 1 x 8$ $color(orange)9 " 1 x 8"$

The signs are the $s a m e$ $color(magenta)(same)$ , they will both be negative.

This gives $(x - 1) (x - 8) = 0$ $(x-1)(x-8) = 0$

Once factored, the equation can be solved to give

$x = 1, x = 8$ $x = 1, x = 8$

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