Solve a^2+3^2=5^2

$$a ^ { 2 } + 3 ^ { 2 } = 5 ^ { 2 }$$

Solve for a

$$a=-4$$

$$a=4$$

Steps Using Difference of Squares

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$a^{2}+9=25$$

Subtract 25 from both sides.

$$a^{2}+9-25=0$$

Subtract 25 from 9 to get -16.

$$a^{2}-16=0$$

Consider a^{2}-16. Rewrite a^{2}-16 as a^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(a-4\right)\left(a+4\right)=0$$

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

$$a=4 a=-4$$

Steps by Finding Square Root

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$a^{2}+9=25$$

Subtract 9 from both sides.

$$a^{2}=25-9$$

Subtract 9 from 25 to get 16.

$$a^{2}=16$$

Take the square root of both sides of the equation.

$$a=4 a=-4$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=5^{2}$$

Calculate 5 to the power of 2 and get 25.

$$a^{2}+9=25$$

Subtract 25 from both sides.

$$a^{2}+9-25=0$$

Subtract 25 from 9 to get -16.

$$a^{2}-16=0$$

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

$$a=\frac{0±\sqrt{0^{2}-4\left(-16\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-16\right)}}{2}$$

Multiply -4 times -16.

$$a=\frac{0±\sqrt{64}}{2}$$

Take the square root of 64.

$$a=\frac{0±8}{2}$$

Now solve the equation a=\frac{0±8}{2} when ± is plus. Divide 8 by 2.

$$a=4$$

Now solve the equation a=\frac{0±8}{2} when ± is minus. Divide -8 by 2.

$$a=-4$$

The equation is now solved.

$$a=4 a=-4$$