How do you solve $x^{2} = x + 6$ $x^2=x+6$ ?

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Answer 1 · 2016-02-04T12:14:44

$x^{2} = x + 6$ $x^2=x+6$

$\to x^{2} - x = 6$ $rarrx^2-x=6$

$\to x^{2} - x - 6 = 0$ $rarrx^2-x-6=0$

Now this is a Quadratic equation (in form $a x^{2} + b x^{2} + c$ $ax^2+bx^2+c$ )

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case $a = 1, b = - 1, c = - 6$ $a=1,b=-1,c=-6$

$\to x = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (- 6)}}{2 (1)}$ $rarrx=(-(-1)+-sqrt((-1)^2-4(1)(-6)))/(2(1))$

$\to x = \frac{1 \pm \sqrt{1 - (- 24)}}{2}$ $rarrx=(1+-sqrt(1-(-24)))/2$

$\to x = \frac{1 \pm \sqrt{1 + 24}}{2}$ $rarrx=(1+-sqrt(1+24))/2$

$\to x = \frac{1 \pm \sqrt{25}}{2}$ $rarrx=(1+-sqrt25)/2$

$\to x = \frac{1 \pm 5}{2}$ $rarrx=(1+-5)/2$

Now we are in a stage in which we take the two answers:

$\to x = \frac{1 + 5}{2}, \frac{1 - 5}{2}$ $rarrx=(1+5)/2,(1-5)/2$

$\to x = \frac{6}{2}, - \frac{4}{2}$ $rarrx=6/2,-4/2$

$\to x = 3, - 2$ $rarrx=3,-2$

Answer 2 · 2016-02-04T12:19:58

This is another way of solving this equation:

$x^{2} = x + 6$ $x^2=x+6$

$\to x^{2} - x - 6 = 0$ $rarrx^2-x-6=0$

Factor:

$\to (x - 3) (x + 2) = 0$ $rarr(x-3)(x+2)=0$

$\to x = 3, - 2$ $rarrx=3,-2$

=)

How do you solve x2=x+6x^2=x+6?