Solve x^2+(x+5)^2=25^2

$$x ^ { 2 } + ( x + 5 ) ^ { 2 } = 25 ^ { 2 }$$

Solve for x

$$x=-20$$

$$x=15$$

Steps Using Factoring By Grouping

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

$$x^{2}+x^{2}+10x+25=25^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+10x+25=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$2x^{2}+10x+25=625$$

Subtract 625 from both sides.

$$2x^{2}+10x+25-625=0$$

Subtract 625 from 25 to get -600.

$$2x^{2}+10x-600=0$$

Divide both sides by 2.

$$x^{2}+5x-300=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-300. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-300\right)=-300$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -300.

$$-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20$$

Calculate the sum for each pair.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 5.

$$a=-15 b=20$$

Rewrite x^{2}+5x-300 as \left(x^{2}-15x\right)+\left(20x-300\right).

$$\left(x^{2}-15x\right)+\left(20x-300\right)$$

Factor out x in the first and 20 in the second group.

$$x\left(x-15\right)+20\left(x-15\right)$$

Factor out common term x-15 by using distributive property.

$$\left(x-15\right)\left(x+20\right)$$

To find equation solutions, solve x-15=0 and x+20=0.

$$x=15 x=-20$$

Steps Using the Quadratic Formula

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

$$x^{2}+x^{2}+10x+25=25^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+10x+25=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$2x^{2}+10x+25=625$$

Subtract 625 from both sides.

$$2x^{2}+10x+25-625=0$$

Subtract 625 from 25 to get -600.

$$2x^{2}+10x-600=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 10 for b, and -600 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\left(-600\right)}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\left(-600\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\left(-600\right)}}{2\times 2}$$

Multiply -8 times -600.

$$x=\frac{-10±\sqrt{100+4800}}{2\times 2}$$

Add 100 to 4800.

$$x=\frac{-10±\sqrt{4900}}{2\times 2}$$

Take the square root of 4900.

$$x=\frac{-10±70}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±70}{4}$$

Now solve the equation x=\frac{-10±70}{4} when ± is plus. Add -10 to 70.

$$x=\frac{60}{4}$$

Divide 60 by 4.

$$x=15$$

Now solve the equation x=\frac{-10±70}{4} when ± is minus. Subtract 70 from -10.

$$x=-\frac{80}{4}$$

Divide -80 by 4.

$$x=-20$$

The equation is now solved.

$$x=15 x=-20$$

Steps for Completing the Square

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

$$x^{2}+x^{2}+10x+25=25^{2}$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+10x+25=25^{2}$$

Calculate 25 to the power of 2 and get 625.

$$2x^{2}+10x+25=625$$

Subtract 25 from both sides.

$$2x^{2}+10x=625-25$$

Subtract 25 from 625 to get 600.

$$2x^{2}+10x=600$$

Divide both sides by 2.

$$\frac{2x^{2}+10x}{2}=\frac{600}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{10}{2}x=\frac{600}{2}$$

Divide 10 by 2.

$$x^{2}+5x=\frac{600}{2}$$

Divide 600 by 2.

$$x^{2}+5x=300$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=300+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=300+\frac{25}{4}$$

Add 300 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{1225}{4}$$

Factor x^{2}+5x+\frac{25}{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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{1225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{1225}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{35}{2} x+\frac{5}{2}=-\frac{35}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=15 x=-20$$