Solve (4x-1)^2=20x-5

$$( 4 x - 1 ) ^ { 2 } = 20 x - 5$$

Solve for x

$$x=\frac{1}{4}=0.25$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

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}.

$$16x^{2}-8x+1=20x-5$$

Subtract 20x from both sides.

$$16x^{2}-8x+1-20x=-5$$

Combine -8x and -20x to get -28x.

$$16x^{2}-28x+1=-5$$

Add 5 to both sides.

$$16x^{2}-28x+1+5=0$$

Add 1 and 5 to get 6.

$$16x^{2}-28x+6=0$$

Divide both sides by 2.

$$8x^{2}-14x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-14 ab=8\times 3=24$$

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 24.

$$-1,-24 -2,-12 -3,-8 -4,-6$$

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -14.

$$a=-12 b=-2$$

Rewrite 8x^{2}-14x+3 as \left(8x^{2}-12x\right)+\left(-2x+3\right).

$$\left(8x^{2}-12x\right)+\left(-2x+3\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(2x-3\right)-\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(4x-1\right)$$

To find equation solutions, solve 2x-3=0 and 4x-1=0.

$$x=\frac{3}{2} x=\frac{1}{4}$$

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}.

$$16x^{2}-8x+1=20x-5$$

Subtract 20x from both sides.

$$16x^{2}-8x+1-20x=-5$$

Combine -8x and -20x to get -28x.

$$16x^{2}-28x+1=-5$$

Add 5 to both sides.

$$16x^{2}-28x+1+5=0$$

Add 1 and 5 to get 6.

$$16x^{2}-28x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -28 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 16\times 6}}{2\times 16}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 16\times 6}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-28\right)±\sqrt{784-64\times 6}}{2\times 16}$$

Multiply -64 times 6.

$$x=\frac{-\left(-28\right)±\sqrt{784-384}}{2\times 16}$$

Add 784 to -384.

$$x=\frac{-\left(-28\right)±\sqrt{400}}{2\times 16}$$

Take the square root of 400.

$$x=\frac{-\left(-28\right)±20}{2\times 16}$$

The opposite of -28 is 28.

$$x=\frac{28±20}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{28±20}{32}$$

Now solve the equation x=\frac{28±20}{32} when ± is plus. Add 28 to 20.

$$x=\frac{48}{32}$$

Reduce the fraction \frac{48}{32} to lowest terms by extracting and canceling out 16.

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

Now solve the equation x=\frac{28±20}{32} when ± is minus. Subtract 20 from 28.

$$x=\frac{8}{32}$$

Reduce the fraction \frac{8}{32} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

$$x=\frac{3}{2} x=\frac{1}{4}$$

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}.

$$16x^{2}-8x+1=20x-5$$

Subtract 20x from both sides.

$$16x^{2}-8x+1-20x=-5$$

Combine -8x and -20x to get -28x.

$$16x^{2}-28x+1=-5$$

Subtract 1 from both sides.

$$16x^{2}-28x=-5-1$$

Subtract 1 from -5 to get -6.

$$16x^{2}-28x=-6$$

Divide both sides by 16.

$$\frac{16x^{2}-28x}{16}=-\frac{6}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{28}{16}\right)x=-\frac{6}{16}$$

Reduce the fraction \frac{-28}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{7}{4}x=-\frac{6}{16}$$

Reduce the fraction \frac{-6}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{7}{4}x=-\frac{3}{8}$$

Divide -\frac{7}{4}, the coefficient of the x term, by 2 to get -\frac{7}{8}. Then add the square of -\frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{4}x+\left(-\frac{7}{8}\right)^{2}=-\frac{3}{8}+\left(-\frac{7}{8}\right)^{2}$$

Square -\frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=-\frac{3}{8}+\frac{49}{64}$$

Add -\frac{3}{8} to \frac{49}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{4}x+\frac{49}{64}=\frac{25}{64}$$

Factor x^{2}-\frac{7}{4}x+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x-\frac{7}{8}=\frac{5}{8} x-\frac{7}{8}=-\frac{5}{8}$$

Add \frac{7}{8} to both sides of the equation.

$$x=\frac{3}{2} x=\frac{1}{4}$$