Solve (x-4)^2-(5-2x)^2=0

$$( x - 4 ) ^ { 2 } - ( 5 - 2 x ) ^ { 2 } = 0$$

Solve for x

$$x=1$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$x^{2}-8x+16-\left(5-2x\right)^{2}=0$$

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

$$x^{2}-8x+16-\left(25-20x+4x^{2}\right)=0$$

To find the opposite of 25-20x+4x^{2}, find the opposite of each term.

$$x^{2}-8x+16-25+20x-4x^{2}=0$$

Subtract 25 from 16 to get -9.

$$x^{2}-8x-9+20x-4x^{2}=0$$

Combine -8x and 20x to get 12x.

$$x^{2}+12x-9-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+12x-9=0$$

Divide both sides by 3.

$$-x^{2}+4x-3=0$$

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

$$a+b=4 ab=-\left(-3\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

Rewrite -x^{2}+4x-3 as \left(-x^{2}+3x\right)+\left(x-3\right).

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

Factor out -x in -x^{2}+3x.

$$-x\left(x-3\right)+x-3$$

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

$$\left(x-3\right)\left(-x+1\right)$$

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

$$x=3 x=1$$

Steps Using the Quadratic Formula

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

$$x^{2}-8x+16-\left(5-2x\right)^{2}=0$$

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

$$x^{2}-8x+16-\left(25-20x+4x^{2}\right)=0$$

To find the opposite of 25-20x+4x^{2}, find the opposite of each term.

$$x^{2}-8x+16-25+20x-4x^{2}=0$$

Subtract 25 from 16 to get -9.

$$x^{2}-8x-9+20x-4x^{2}=0$$

Combine -8x and 20x to get 12x.

$$x^{2}+12x-9-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+12x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 12 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-3\right)\left(-9\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-12±\sqrt{144+12\left(-9\right)}}{2\left(-3\right)}$$

Multiply 12 times -9.

$$x=\frac{-12±\sqrt{144-108}}{2\left(-3\right)}$$

Add 144 to -108.

$$x=\frac{-12±\sqrt{36}}{2\left(-3\right)}$$

Take the square root of 36.

$$x=\frac{-12±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-12±6}{-6}$$

Now solve the equation x=\frac{-12±6}{-6} when ± is plus. Add -12 to 6.

$$x=-\frac{6}{-6}$$

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-12±6}{-6} when ± is minus. Subtract 6 from -12.

$$x=-\frac{18}{-6}$$

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=1 x=3$$

Steps for Completing the Square

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

$$x^{2}-8x+16-\left(5-2x\right)^{2}=0$$

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

$$x^{2}-8x+16-\left(25-20x+4x^{2}\right)=0$$

To find the opposite of 25-20x+4x^{2}, find the opposite of each term.

$$x^{2}-8x+16-25+20x-4x^{2}=0$$

Subtract 25 from 16 to get -9.

$$x^{2}-8x-9+20x-4x^{2}=0$$

Combine -8x and 20x to get 12x.

$$x^{2}+12x-9-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+12x-9=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$-3x^{2}+12x=9$$

Divide both sides by -3.

$$\frac{-3x^{2}+12x}{-3}=\frac{9}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{12}{-3}x=\frac{9}{-3}$$

Divide 12 by -3.

$$x^{2}-4x=\frac{9}{-3}$$

Divide 9 by -3.

$$x^{2}-4x=-3$$

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

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

Square -2.

$$x^{2}-4x+4=-3+4$$

Add -3 to 4.

$$x^{2}-4x+4=1$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=1 x-2=-1$$

Add 2 to both sides of the equation.

$$x=3 x=1$$