Solve 9(x-1)^2-4=0

$$9 ( x - 1 ) ^ { 2 } - 4 = 0$$

Solve for x

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$9\left(x^{2}-2x+1\right)-4=0$$

Use the distributive property to multiply 9 by x^{2}-2x+1.

$$9x^{2}-18x+9-4=0$$

Subtract 4 from 9 to get 5.

$$9x^{2}-18x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-18 ab=9\times 5=45$$

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 45.

$$-1,-45 -3,-15 -5,-9$$

Calculate the sum for each pair.

$$-1-45=-46 -3-15=-18 -5-9=-14$$

The solution is the pair that gives sum -18.

$$a=-15 b=-3$$

Rewrite 9x^{2}-18x+5 as \left(9x^{2}-15x\right)+\left(-3x+5\right).

$$\left(9x^{2}-15x\right)+\left(-3x+5\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(3x-5\right)-\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(3x-1\right)$$

To find equation solutions, solve 3x-5=0 and 3x-1=0.

$$x=\frac{5}{3} x=\frac{1}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$9\left(x^{2}-2x+1\right)-4=0$$

Use the distributive property to multiply 9 by x^{2}-2x+1.

$$9x^{2}-18x+9-4=0$$

Subtract 4 from 9 to get 5.

$$9x^{2}-18x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -18 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 9\times 5}}{2\times 9}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 9\times 5}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-18\right)±\sqrt{324-36\times 5}}{2\times 9}$$

Multiply -36 times 5.

$$x=\frac{-\left(-18\right)±\sqrt{324-180}}{2\times 9}$$

Add 324 to -180.

$$x=\frac{-\left(-18\right)±\sqrt{144}}{2\times 9}$$

Take the square root of 144.

$$x=\frac{-\left(-18\right)±12}{2\times 9}$$

The opposite of -18 is 18.

$$x=\frac{18±12}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{18±12}{18}$$

Now solve the equation x=\frac{18±12}{18} when ± is plus. Add 18 to 12.

$$x=\frac{30}{18}$$

Reduce the fraction \frac{30}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{5}{3}$$

Now solve the equation x=\frac{18±12}{18} when ± is minus. Subtract 12 from 18.

$$x=\frac{6}{18}$$

Reduce the fraction \frac{6}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{1}{3}$$

The equation is now solved.

$$x=\frac{5}{3} x=\frac{1}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$9\left(x^{2}-2x+1\right)-4=0$$

Use the distributive property to multiply 9 by x^{2}-2x+1.

$$9x^{2}-18x+9-4=0$$

Subtract 4 from 9 to get 5.

$$9x^{2}-18x+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-18x=-5$$

Divide both sides by 9.

$$\frac{9x^{2}-18x}{9}=-\frac{5}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{18}{9}\right)x=-\frac{5}{9}$$

Divide -18 by 9.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=-\frac{5}{9}+1$$

Add -\frac{5}{9} to 1.

$$x^{2}-2x+1=\frac{4}{9}$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-1=\frac{2}{3} x-1=-\frac{2}{3}$$

Add 1 to both sides of the equation.

$$x=\frac{5}{3} x=\frac{1}{3}$$