How do you solve $x² - 4x - 8 = 4$ ?

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Answer 1 · 2016-04-09T07:37:51

$x=6,-2$

$color(blue)(x^2-4x-8=4$

Subtract $4$ both sides

$rarrx^2-4x-8-4=4-4$

$rarrx^2-4x-12=0$

This is a Quadratic equation (in form $ax^2+bx+c=0$ )

$color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)$

Remember that, $aandb$ are the coefficients and $c$ is the constant term

So,

$color(orange)(a=1,b=-4,c=-12$

$rarrx=(-(-4)+-sqrt(-4^2-4(1)(-12)))/(2(1))$

$rarrx=(4+-sqrt(16-(-48)))/(2)$

$rarrx=(4+-sqrt(16+48))/(2)$

$rarrx=(4+-sqrt(64))/(2)$

$rarrx=(4+-8)/(2)$

Now we have two solutions

$color(purple)(x=(4+8)/(2)=12/2=6$

$color(violet)(x=(4-8)/(2)=-4/2=-2$

How do you solve x² - 4x - 8 = 4x² - 4x - 8 = 4?