Solve (5x-4)^2=81

$$( 5 x - 4 ) ^ { 2 } = 81$$

Solve for x

$$x = \frac{13}{5} = 2\frac{3}{5} = 2.6$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16=81$$

Subtract 81 from both sides.

$$25x^{2}-40x+16-81=0$$

Subtract 81 from 16 to get -65.

$$25x^{2}-40x-65=0$$

Divide both sides by 5.

$$5x^{2}-8x-13=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-13. To find a and b, set up a system to be solved.

$$a+b=-8 ab=5\left(-13\right)=-65$$

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 -65.

$$1,-65 5,-13$$

Calculate the sum for each pair.

$$1-65=-64 5-13=-8$$

The solution is the pair that gives sum -8.

$$a=-13 b=5$$

Rewrite 5x^{2}-8x-13 as \left(5x^{2}-13x\right)+\left(5x-13\right).

$$\left(5x^{2}-13x\right)+\left(5x-13\right)$$

Factor out x in 5x^{2}-13x.

$$x\left(5x-13\right)+5x-13$$

Factor out common term 5x-13 by using distributive property.

$$\left(5x-13\right)\left(x+1\right)$$

To find equation solutions, solve 5x-13=0 and x+1=0.

$$x=\frac{13}{5} x=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16=81$$

Subtract 81 from both sides.

$$25x^{2}-40x+16-81=0$$

Subtract 81 from 16 to get -65.

$$25x^{2}-40x-65=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -40 for b, and -65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 25\left(-65\right)}}{2\times 25}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 25\left(-65\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-40\right)±\sqrt{1600-100\left(-65\right)}}{2\times 25}$$

Multiply -100 times -65.

$$x=\frac{-\left(-40\right)±\sqrt{1600+6500}}{2\times 25}$$

Add 1600 to 6500.

$$x=\frac{-\left(-40\right)±\sqrt{8100}}{2\times 25}$$

Take the square root of 8100.

$$x=\frac{-\left(-40\right)±90}{2\times 25}$$

The opposite of -40 is 40.

$$x=\frac{40±90}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{40±90}{50}$$

Now solve the equation x=\frac{40±90}{50} when ± is plus. Add 40 to 90.

$$x=\frac{130}{50}$$

Reduce the fraction \frac{130}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{13}{5}$$

Now solve the equation x=\frac{40±90}{50} when ± is minus. Subtract 90 from 40.

$$x=-\frac{50}{50}$$

Divide -50 by 50.

$$x=-1$$

The equation is now solved.

$$x=\frac{13}{5} x=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-4\right)^{2}.

$$25x^{2}-40x+16=81$$

Subtract 16 from both sides.

$$25x^{2}-40x=81-16$$

Subtract 16 from 81 to get 65.

$$25x^{2}-40x=65$$

Divide both sides by 25.

$$\frac{25x^{2}-40x}{25}=\frac{65}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{40}{25}\right)x=\frac{65}{25}$$

Reduce the fraction \frac{-40}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{8}{5}x=\frac{65}{25}$$

Reduce the fraction \frac{65}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{8}{5}x=\frac{13}{5}$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=\frac{13}{5}+\left(-\frac{4}{5}\right)^{2}$$

Square -\frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{13}{5}+\frac{16}{25}$$

Add \frac{13}{5} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{81}{25}$$

Factor x^{2}-\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{5}\right)^{2}=\frac{81}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{\frac{81}{25}}$$

Simplify.

$$x-\frac{4}{5}=\frac{9}{5} x-\frac{4}{5}=-\frac{9}{5}$$

Add \frac{4}{5} to both sides of the equation.

$$x=\frac{13}{5} x=-1$$