Solve 9x^2-(5x-1)^2=2x

$$9 x ^ { 2 } - ( 5 x - 1 ) ^ { 2 } = 2 x$$

Solve for x

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

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}.

$$9x^{2}-\left(25x^{2}-10x+1\right)=2x$$

To find the opposite of 25x^{2}-10x+1, find the opposite of each term.

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

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

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

Subtract 2x from both sides.

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

Combine 10x and -2x to get 8x.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -16x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=8 ab=-16\left(-1\right)=16$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 16.

$$1,16 2,8 4,4$$

Calculate the sum for each pair.

$$1+16=17 2+8=10 4+4=8$$

The solution is the pair that gives sum 8.

$$a=4 b=4$$

Rewrite -16x^{2}+8x-1 as \left(-16x^{2}+4x\right)+\left(4x-1\right).

$$\left(-16x^{2}+4x\right)+\left(4x-1\right)$$

Factor out -4x in -16x^{2}+4x.

$$-4x\left(4x-1\right)+4x-1$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(-4x+1\right)$$

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

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

Steps Using the Quadratic Formula

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

$$9x^{2}-\left(25x^{2}-10x+1\right)=2x$$

To find the opposite of 25x^{2}-10x+1, find the opposite of each term.

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

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

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

Subtract 2x from both sides.

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

Combine 10x and -2x to get 8x.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 8 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-16\right)\left(-1\right)}}{2\left(-16\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-16\right)\left(-1\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-8±\sqrt{64+64\left(-1\right)}}{2\left(-16\right)}$$

Multiply 64 times -1.

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

Add 64 to -64.

$$x=\frac{-8±\sqrt{0}}{2\left(-16\right)}$$

Take the square root of 0.

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

Multiply 2 times -16.

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

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

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

Steps for Completing the Square

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

$$9x^{2}-\left(25x^{2}-10x+1\right)=2x$$

To find the opposite of 25x^{2}-10x+1, find the opposite of each term.

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

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

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

Subtract 2x from both sides.

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

Combine 10x and -2x to get 8x.

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

Add 1 to both sides. Anything plus zero gives itself.

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

Divide both sides by -16.

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

Dividing by -16 undoes the multiplication by -16.

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

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

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

Divide 1 by -16.

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

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.

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

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

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

Add -\frac{1}{16} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=0$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{1}{4}=0 x-\frac{1}{4}=0$$

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

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

The equation is now solved. Solutions are the same.

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