Solve (4x-1)^2=225

$$( 4 x - 1 ) ^ { 2 } = 225$$

Solve for x

$$x=4$$

$$x = -\frac{7}{2} = -3\frac{1}{2} = -3.5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-1\right)^{2}.

$$16x^{2}-8x+1=225$$

Subtract 225 from both sides.

$$16x^{2}-8x+1-225=0$$

Subtract 225 from 1 to get -224.

$$16x^{2}-8x-224=0$$

Divide both sides by 8.

$$2x^{2}-x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-28\right)=-56$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -56.

$$1,-56 2,-28 4,-14 7,-8$$

Calculate the sum for each pair.

$$1-56=-55 2-28=-26 4-14=-10 7-8=-1$$

The solution is the pair that gives sum -1.

$$a=-8 b=7$$

Rewrite 2x^{2}-x-28 as \left(2x^{2}-8x\right)+\left(7x-28\right).

$$\left(2x^{2}-8x\right)+\left(7x-28\right)$$

Factor out 2x in the first and 7 in the second group.

$$2x\left(x-4\right)+7\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(2x+7\right)$$

To find equation solutions, solve x-4=0 and 2x+7=0.

$$x=4 x=-\frac{7}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-1\right)^{2}.

$$16x^{2}-8x+1=225$$

Subtract 225 from both sides.

$$16x^{2}-8x+1-225=0$$

Subtract 225 from 1 to get -224.

$$16x^{2}-8x-224=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -8 for b, and -224 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 16\left(-224\right)}}{2\times 16}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 16\left(-224\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-8\right)±\sqrt{64-64\left(-224\right)}}{2\times 16}$$

Multiply -64 times -224.

$$x=\frac{-\left(-8\right)±\sqrt{64+14336}}{2\times 16}$$

Add 64 to 14336.

$$x=\frac{-\left(-8\right)±\sqrt{14400}}{2\times 16}$$

Take the square root of 14400.

$$x=\frac{-\left(-8\right)±120}{2\times 16}$$

The opposite of -8 is 8.

$$x=\frac{8±120}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{8±120}{32}$$

Now solve the equation x=\frac{8±120}{32} when ± is plus. Add 8 to 120.

$$x=\frac{128}{32}$$

Divide 128 by 32.

$$x=4$$

Now solve the equation x=\frac{8±120}{32} when ± is minus. Subtract 120 from 8.

$$x=-\frac{112}{32}$$

Reduce the fraction \frac{-112}{32} to lowest terms by extracting and canceling out 16.

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

The equation is now solved.

$$x=4 x=-\frac{7}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x-1\right)^{2}.

$$16x^{2}-8x+1=225$$

Subtract 1 from both sides.

$$16x^{2}-8x=225-1$$

Subtract 1 from 225 to get 224.

$$16x^{2}-8x=224$$

Divide both sides by 16.

$$\frac{16x^{2}-8x}{16}=\frac{224}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{8}{16}\right)x=\frac{224}{16}$$

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{1}{2}x=\frac{224}{16}$$

Divide 224 by 16.

$$x^{2}-\frac{1}{2}x=14$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=14+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=14+\frac{1}{16}$$

Add 14 to \frac{1}{16}.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{225}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{15}{4} x-\frac{1}{4}=-\frac{15}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=4 x=-\frac{7}{2}$$