Solve a^2+3^2=109

$$a ^ { 2 } + 3 ^ { 2 } = 109$$

Solve for a

$$a=-10$$

$$a=10$$

Steps Using Difference of Squares

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=109$$

Subtract 109 from both sides.

$$a^{2}+9-109=0$$

Subtract 109 from 9 to get -100.

$$a^{2}-100=0$$

Consider a^{2}-100. Rewrite a^{2}-100 as a^{2}-10^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(a-10\right)\left(a+10\right)=0$$

To find equation solutions, solve a-10=0 and a+10=0.

$$a=10 a=-10$$

Steps by Finding Square Root

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=109$$

Subtract 9 from both sides.

$$a^{2}=109-9$$

Subtract 9 from 109 to get 100.

$$a^{2}=100$$

Take the square root of both sides of the equation.

$$a=10 a=-10$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$a^{2}+9=109$$

Subtract 109 from both sides.

$$a^{2}+9-109=0$$

Subtract 109 from 9 to get -100.

$$a^{2}-100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{0±\sqrt{0^{2}-4\left(-100\right)}}{2}$$

Square 0.

$$a=\frac{0±\sqrt{-4\left(-100\right)}}{2}$$

Multiply -4 times -100.

$$a=\frac{0±\sqrt{400}}{2}$$

Take the square root of 400.

$$a=\frac{0±20}{2}$$

Now solve the equation a=\frac{0±20}{2} when ± is plus. Divide 20 by 2.

$$a=10$$

Now solve the equation a=\frac{0±20}{2} when ± is minus. Divide -20 by 2.

$$a=-10$$

The equation is now solved.

$$a=10 a=-10$$