Solve 8x=x^2-9

$$8x= { x }^{ 2 } -9$$

Solve for x

$$x=-1$$

$$x=9$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$8x-x^{2}=-9$$

Add 9 to both sides.

$$8x-x^{2}+9=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+8x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=8 ab=-9=-9$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -9.

$$-1,9 -3,3$$

Calculate the sum for each pair.

$$-1+9=8 -3+3=0$$

The solution is the pair that gives sum 8.

$$a=9 b=-1$$

Rewrite -x^{2}+8x+9 as \left(-x^{2}+9x\right)+\left(-x+9\right).

$$\left(-x^{2}+9x\right)+\left(-x+9\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(x-9\right)-\left(x-9\right)$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x-1\right)$$

To find equation solutions, solve x-9=0 and -x-1=0.

$$x=9 x=-1$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$8x-x^{2}=-9$$

Add 9 to both sides.

$$8x-x^{2}+9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+8x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 8 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-1\right)\times 9}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-8±\sqrt{64+4\times 9}}{2\left(-1\right)}$$

Multiply 4 times 9.

$$x=\frac{-8±\sqrt{64+36}}{2\left(-1\right)}$$

Add 64 to 36.

$$x=\frac{-8±\sqrt{100}}{2\left(-1\right)}$$

Take the square root of 100.

$$x=\frac{-8±10}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-8±10}{-2}$$

Now solve the equation x=\frac{-8±10}{-2} when ± is plus. Add -8 to 10.

$$x=\frac{2}{-2}$$

Divide 2 by -2.

$$x=-1$$

Now solve the equation x=\frac{-8±10}{-2} when ± is minus. Subtract 10 from -8.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

The equation is now solved.

$$x=-1 x=9$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$8x-x^{2}=-9$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+8x=-9$$

Divide both sides by -1.

$$\frac{-x^{2}+8x}{-1}=-\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{8}{-1}x=-\frac{9}{-1}$$

Divide 8 by -1.

$$x^{2}-8x=-\frac{9}{-1}$$

Divide -9 by -1.

$$x^{2}-8x=9$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=9+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=9+16$$

Add 9 to 16.

$$x^{2}-8x+16=25$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-4=5 x-4=-5$$

Add 4 to both sides of the equation.

$$x=9 x=-1$$