How do you solve #x + 4 = 3x²#?

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Answer 1 · 2016-03-24T12:51:45

#x=4/3,-1#

#color(blue)(x+4=3x^2#

#rarrx=3x^2-4#

#rarr0=3x^2-4-x#

Rewrite in standard form

#color(purple)(rarr3x^2-x-4=0#

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

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

Where

#color(red)(a=3,b=-1,c=-4#

#rarrx=(-(-1)+-sqrt(-1^2-4(3)(-4)))/(2(3))#

#rarrx=(1+-sqrt(1-4(-12)))/(6)#

#rarrx=(1+-sqrt(1-(-48)))/(6)#

#rarrx=(1+-sqrt(49))/(6)#

#rarrx=(1+-7)/(6)#

Now we have #2# values for #x#

#color(indigo)( x=(1+7)/(6)=8/6=4/3#

#color(orange)(x=(1-7)/(6)=-6/6=-1#

#color(blue)(ul bar |x=4/3,-1|#