Solve 5^2-x^2=(10-2x)^2

$$5 ^ { 2 } - x ^ { 2 } = ( 10 - 2 x ) ^ { 2 }$$

Solve for x

$$x=3$$

$$x=5$$

Steps Using Factoring By Grouping

Calculate 5 to the power of 2 and get 25.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-2x\right)^{2}.

$$25-x^{2}=100-40x+4x^{2}$$

Subtract 100 from both sides.

$$25-x^{2}-100=-40x+4x^{2}$$

Subtract 100 from 25 to get -75.

$$-75-x^{2}=-40x+4x^{2}$$

Add 40x to both sides.

$$-75-x^{2}+40x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$-75-x^{2}+40x-4x^{2}=0$$

Combine -x^{2} and -4x^{2} to get -5x^{2}.

$$-75-5x^{2}+40x=0$$

Divide both sides by 5.

$$-15-x^{2}+8x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+8x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=8 ab=-\left(-15\right)=15$$

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 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 8.

$$a=5 b=3$$

Rewrite -x^{2}+8x-15 as \left(-x^{2}+5x\right)+\left(3x-15\right).

$$\left(-x^{2}+5x\right)+\left(3x-15\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-5\right)+3\left(x-5\right)$$

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

$$\left(x-5\right)\left(-x+3\right)$$

To find equation solutions, solve x-5=0 and -x+3=0.

$$x=5 x=3$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-2x\right)^{2}.

$$25-x^{2}=100-40x+4x^{2}$$

Subtract 100 from both sides.

$$25-x^{2}-100=-40x+4x^{2}$$

Subtract 100 from 25 to get -75.

$$-75-x^{2}=-40x+4x^{2}$$

Add 40x to both sides.

$$-75-x^{2}+40x=4x^{2}$$

Subtract 4x^{2} from both sides.

$$-75-x^{2}+40x-4x^{2}=0$$

Combine -x^{2} and -4x^{2} to get -5x^{2}.

$$-75-5x^{2}+40x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-5x^{2}+40x-75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 40 for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-5\right)\left(-75\right)}}{2\left(-5\right)}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-5\right)\left(-75\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-40±\sqrt{1600+20\left(-75\right)}}{2\left(-5\right)}$$

Multiply 20 times -75.

$$x=\frac{-40±\sqrt{1600-1500}}{2\left(-5\right)}$$

Add 1600 to -1500.

$$x=\frac{-40±\sqrt{100}}{2\left(-5\right)}$$

Take the square root of 100.

$$x=\frac{-40±10}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-40±10}{-10}$$

Now solve the equation x=\frac{-40±10}{-10} when ± is plus. Add -40 to 10.

$$x=-\frac{30}{-10}$$

Divide -30 by -10.

$$x=3$$

Now solve the equation x=\frac{-40±10}{-10} when ± is minus. Subtract 10 from -40.

$$x=-\frac{50}{-10}$$

Divide -50 by -10.

$$x=5$$

The equation is now solved.

$$x=3 x=5$$

Steps for Completing the Square

Calculate 5 to the power of 2 and get 25.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-2x\right)^{2}.

$$25-x^{2}=100-40x+4x^{2}$$

Add 40x to both sides.

$$25-x^{2}+40x=100+4x^{2}$$

Subtract 4x^{2} from both sides.

$$25-x^{2}+40x-4x^{2}=100$$

Combine -x^{2} and -4x^{2} to get -5x^{2}.

$$25-5x^{2}+40x=100$$

Subtract 25 from both sides.

$$-5x^{2}+40x=100-25$$

Subtract 25 from 100 to get 75.

$$-5x^{2}+40x=75$$

Divide both sides by -5.

$$\frac{-5x^{2}+40x}{-5}=\frac{75}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{40}{-5}x=\frac{75}{-5}$$

Divide 40 by -5.

$$x^{2}-8x=\frac{75}{-5}$$

Divide 75 by -5.

$$x^{2}-8x=-15$$

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

$$x^{2}-8x+\left(-4\right)^{2}=-15+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=-15+16$$

Add -15 to 16.

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

Factor x^{2}-8x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=1 x-4=-1$$

Add 4 to both sides of the equation.

$$x=5 x=3$$