How do you solve $3 x^{2} = x + 4$ $3x^2=x+4$ by factoring?

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Answer 1 · 2016-10-26T06:28:06

$x = 1 \frac{1}{3}$ $x=1 1/3$ , $- 1$ $-1$

$3 x^{2} = x + 4$ $3x^2=x+4$

Shift all the terms to the left.

$3 x^{2} - x - 4 = 0$ $3x^2-x-4=0$

Factorise.

$3 x^{2} + 3 x - 4 x - 4 = 0$ $3x^2+3x-4x-4=0$

$3 x (x + 1) - 4 (x + 1) = 0$ $3x(x+1)-4(x+1)=0$

$(3 x - 4) (x + 1) = 0$ $(3x-4)(x+1)=0$

$3 x - 4 = 0$ $3x-4=0$ and $x + 1 = 0$ $x+1=0$

$x = \frac{4}{3} = 1 \frac{1}{3}$ $x=4/3 = 1 1/3$ and $x = - 1$ $x=-1$

How do you solve 3x2=x+43x^2=x+4 by factoring?