How do you solve $x^{2} - x = 5 x - 9$ $x^2 -x = 5x - 9$ ?

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Answer 1 · 2015-07-10T09:03:43

$x = 3$ $x=3$

$x^{2} - x = 5 x - 9$ $x^2-x=5x-9$

Subtract $5 x$ $5x$ from both sides of the equation.

$x^{2} - x - 5 x = - 9$ $x^2-x-5x=-9$ =

$x^{2} - 6 x = - 9$ $x^2-6x=-9$

Add $9$ $9$ to both sides.

$x^{2} - 6 x + 9 = 0$ $x^2-6x+9=0$

We now have a quadratic equation with the form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , where $a = 1;$ $a=1;$ $b = - 6;$ $b=-6;$ and $c = 9$ $c=9$ .

We can solve for $x$ $x$ using the quadratic formula.

Quadratic Formula

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

$x = \frac{- (- 6) \pm \sqrt{- 6^{2} - 4 \cdot 1 \cdot 9}}{2 \cdot 1}$ $x=(-(-6)+-sqrt(-6^2-4*1*9))/(2*1)$ =

$x = \frac{6 \pm \sqrt{36 - 36}}{2}$ $x=(6+-sqrt(36-36))/2$ =

$x = \frac{6 \pm \sqrt{0}}{2}$ $x=(6+-sqrt0)/2$ =

$x = \frac{6 \pm 0}{2}$ $x=(6+-0)/2$ =

$x = \frac{6}{2}$ $x=6/2$ =

$x = 3$ $x=3$

How do you solve x2−x=5x−9x^2 -x = 5x - 9?