Solve 8x^2=9x-1

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

Solve for x

$$x=\frac{1}{8}=0.125$$

$$x=1$$

Steps Using Factoring By Grouping

Subtract 9x from both sides.

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

Add 1 to both sides.

$$8x^{2}-9x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

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

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 8x^{2}-9x+1 as \left(8x^{2}-8x\right)+\left(-x+1\right).

$$\left(8x^{2}-8x\right)+\left(-x+1\right)$$

Factor out 8x in the first and -1 in the second group.

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

Factor out common term x-1 by using distributive property.

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

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

$$x=1 x=\frac{1}{8}$$

Steps Using the Quadratic Formula

Subtract 9x from both sides.

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

Add 1 to both sides.

$$8x^{2}-9x+1=0$$

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

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

Square -9.

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

Multiply -4 times 8.

$$x=\frac{-\left(-9\right)±\sqrt{81-32}}{2\times 8}$$

Add 81 to -32.

$$x=\frac{-\left(-9\right)±\sqrt{49}}{2\times 8}$$

Take the square root of 49.

$$x=\frac{-\left(-9\right)±7}{2\times 8}$$

The opposite of -9 is 9.

$$x=\frac{9±7}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{9±7}{16}$$

Now solve the equation x=\frac{9±7}{16} when ± is plus. Add 9 to 7.

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

Divide 16 by 16.

$$x=1$$

Now solve the equation x=\frac{9±7}{16} when ± is minus. Subtract 7 from 9.

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

Reduce the fraction \frac{2}{16} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{8}$$

The equation is now solved.

$$x=1 x=\frac{1}{8}$$

Steps for Completing the Square

Subtract 9x from both sides.

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

Divide both sides by 8.

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

Dividing by 8 undoes the multiplication by 8.

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

Divide -\frac{9}{8}, the coefficient of the x term, by 2 to get -\frac{9}{16}. Then add the square of -\frac{9}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{8}x+\left(-\frac{9}{16}\right)^{2}=-\frac{1}{8}+\left(-\frac{9}{16}\right)^{2}$$

Square -\frac{9}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{8}x+\frac{81}{256}=-\frac{1}{8}+\frac{81}{256}$$

Add -\frac{1}{8} to \frac{81}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{8}x+\frac{81}{256}=\frac{49}{256}$$

Factor x^{2}-\frac{9}{8}x+\frac{81}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{16}\right)^{2}=\frac{49}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{16}\right)^{2}}=\sqrt{\frac{49}{256}}$$

Simplify.

$$x-\frac{9}{16}=\frac{7}{16} x-\frac{9}{16}=-\frac{7}{16}$$

Add \frac{9}{16} to both sides of the equation.

$$x=1 x=\frac{1}{8}$$