Solve 12^2+x^2=13^2

$$12 ^ { 2 } + x ^ { 2 } = 13 ^ { 2 }$$

Solve for x

$$x=-5$$

$$x=5$$

Steps Using Difference of Squares

Calculate 12 to the power of 2 and get 144.

$$144+x^{2}=13^{2}$$

Calculate 13 to the power of 2 and get 169.

$$144+x^{2}=169$$

Subtract 169 from both sides.

$$144+x^{2}-169=0$$

Subtract 169 from 144 to get -25.

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

Consider -25+x^{2}. Rewrite -25+x^{2} as x^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-5\right)\left(x+5\right)=0$$

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

$$x=5 x=-5$$

Steps by Finding Square Root

Calculate 12 to the power of 2 and get 144.

$$144+x^{2}=13^{2}$$

Calculate 13 to the power of 2 and get 169.

$$144+x^{2}=169$$

Subtract 144 from both sides.

$$x^{2}=169-144$$

Subtract 144 from 169 to get 25.

$$x^{2}=25$$

Take the square root of both sides of the equation.

$$x=5 x=-5$$

Steps Using the Quadratic Formula

Calculate 12 to the power of 2 and get 144.

$$144+x^{2}=13^{2}$$

Calculate 13 to the power of 2 and get 169.

$$144+x^{2}=169$$

Subtract 169 from both sides.

$$144+x^{2}-169=0$$

Subtract 169 from 144 to get -25.

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

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$x^{2}-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-25\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-25\right)}}{2}$$

Multiply -4 times -25.

$$x=\frac{0±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{0±10}{2}$$

Now solve the equation x=\frac{0±10}{2} when ± is plus. Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{0±10}{2} when ± is minus. Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=5 x=-5$$