Solve 3^2+b^2=18

$$3 ^ { 2 } + b ^ { 2 } = 18$$

Solve for b

$$b=-3$$

$$b=3$$

Steps Using Difference of Squares

Calculate 3 to the power of 2 and get 9.

$$9+b^{2}=18$$

Subtract 18 from both sides.

$$9+b^{2}-18=0$$

Subtract 18 from 9 to get -9.

$$-9+b^{2}=0$$

Consider -9+b^{2}. Rewrite -9+b^{2} as b^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(b-3\right)\left(b+3\right)=0$$

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

$$b=3 b=-3$$

Steps by Finding Square Root

Calculate 3 to the power of 2 and get 9.

$$9+b^{2}=18$$

Subtract 9 from both sides.

$$b^{2}=18-9$$

Subtract 9 from 18 to get 9.

$$b^{2}=9$$

Take the square root of both sides of the equation.

$$b=3 b=-3$$

Steps Using the Quadratic Formula

Calculate 3 to the power of 2 and get 9.

$$9+b^{2}=18$$

Subtract 18 from both sides.

$$9+b^{2}-18=0$$

Subtract 18 from 9 to get -9.

$$-9+b^{2}=0$$

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.

$$b^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$

Square 0.

$$b=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$b=\frac{0±\sqrt{36}}{2}$$

Take the square root of 36.

$$b=\frac{0±6}{2}$$

Now solve the equation b=\frac{0±6}{2} when ± is plus. Divide 6 by 2.

$$b=3$$

Now solve the equation b=\frac{0±6}{2} when ± is minus. Divide -6 by 2.

$$b=-3$$

The equation is now solved.

$$b=3 b=-3$$