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 $x$

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

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

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

How do you solve x + 4 = 3x²x + 4 = 3x²?