Solve (5x-1)^2-16x^2=0

$$( 5 x - 1 ) ^ { 2 } - 16 x ^ { 2 } = 0$$

Solve for x

$$x=\frac{1}{9}\approx 0.111111111$$

$$x=1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1-16x^{2}=0$$

Combine 25x^{2} and -16x^{2} to get 9x^{2}.

$$9x^{2}-10x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-10 ab=9\times 1=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -10.

$$a=-9 b=-1$$

Rewrite 9x^{2}-10x+1 as \left(9x^{2}-9x\right)+\left(-x+1\right).

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

Factor out 9x in the first and -1 in the second group.

$$9x\left(x-1\right)-\left(x-1\right)$$

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

$$\left(x-1\right)\left(9x-1\right)$$

To find equation solutions, solve x-1=0 and 9x-1=0.

$$x=1 x=\frac{1}{9}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1-16x^{2}=0$$

Combine 25x^{2} and -16x^{2} to get 9x^{2}.

$$9x^{2}-10x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -10 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 9}}{2\times 9}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 9}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-10\right)±\sqrt{100-36}}{2\times 9}$$

Add 100 to -36.

$$x=\frac{-\left(-10\right)±\sqrt{64}}{2\times 9}$$

Take the square root of 64.

$$x=\frac{-\left(-10\right)±8}{2\times 9}$$

The opposite of -10 is 10.

$$x=\frac{10±8}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{10±8}{18}$$

Now solve the equation x=\frac{10±8}{18} when ± is plus. Add 10 to 8.

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

Divide 18 by 18.

$$x=1$$

Now solve the equation x=\frac{10±8}{18} when ± is minus. Subtract 8 from 10.

$$x=\frac{2}{18}$$

Reduce the fraction \frac{2}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{9}$$

The equation is now solved.

$$x=1 x=\frac{1}{9}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-1\right)^{2}.

$$25x^{2}-10x+1-16x^{2}=0$$

Combine 25x^{2} and -16x^{2} to get 9x^{2}.

$$9x^{2}-10x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-10x=-1$$

Divide both sides by 9.

$$\frac{9x^{2}-10x}{9}=-\frac{1}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{10}{9}x=-\frac{1}{9}$$

Divide -\frac{10}{9}, the coefficient of the x term, by 2 to get -\frac{5}{9}. Then add the square of -\frac{5}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{9}x+\left(-\frac{5}{9}\right)^{2}=-\frac{1}{9}+\left(-\frac{5}{9}\right)^{2}$$

Square -\frac{5}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=-\frac{1}{9}+\frac{25}{81}$$

Add -\frac{1}{9} to \frac{25}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{10}{9}x+\frac{25}{81}=\frac{16}{81}$$

Factor x^{2}-\frac{10}{9}x+\frac{25}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{9}\right)^{2}=\frac{16}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{9}\right)^{2}}=\sqrt{\frac{16}{81}}$$

Simplify.

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

Add \frac{5}{9} to both sides of the equation.

$$x=1 x=\frac{1}{9}$$