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#