Solve x^2=9x-8

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

Solve for x

$$x=1$$

$$x=8$$

Steps Using Factoring

Subtract 9x from both sides.

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

Add 8 to both sides.

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

To solve the equation, factor x^{2}-9x+8 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

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

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -9.

$$a=-8 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-8\right)\left(x-1\right)$$

To find equation solutions, solve x-8=0 and x-1=0.

$$x=8 x=1$$

Steps Using Factoring By Grouping

Subtract 9x from both sides.

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

Add 8 to both sides.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -9.

$$a=-8 b=-1$$

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

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

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

$$x\left(x-8\right)-\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-1\right)$$

To find equation solutions, solve x-8=0 and x-1=0.

$$x=8 x=1$$

Steps Using the Quadratic Formula

Subtract 9x from both sides.

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

Add 8 to both sides.

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

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

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

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times 8}}{2}$$

Multiply -4 times 8.

$$x=\frac{-\left(-9\right)±\sqrt{81-32}}{2}$$

Add 81 to -32.

$$x=\frac{-\left(-9\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$x=\frac{-\left(-9\right)±7}{2}$$

The opposite of -9 is 9.

$$x=\frac{9±7}{2}$$

Now solve the equation x=\frac{9±7}{2} when ± is plus. Add 9 to 7.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{9±7}{2} when ± is minus. Subtract 7 from 9.

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

Divide 2 by 2.

$$x=1$$

The equation is now solved.

$$x=8 x=1$$

Steps for Completing the Square

Subtract 9x from both sides.

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

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-8+\frac{81}{4}$$

Add -8 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{49}{4}$$

Factor x^{2}-9x+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{7}{2} x-\frac{9}{2}=-\frac{7}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=8 x=1$$