Solve 4^2+a^2=(3sqrt{2})^2

$$4 ^ { 2 } + a ^ { 2 } = ( 3 \sqrt { 2 } ) ^ { 2 }$$

Solve for a

$$a=\sqrt{2}\approx 1.414213562$$

$$a=-\sqrt{2}\approx -1.414213562$$

Steps by Finding Square Root

Calculate 4 to the power of 2 and get 16.

$$16+a^{2}=\left(3\sqrt{2}\right)^{2}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$16+a^{2}=3^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$16+a^{2}=9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$16+a^{2}=9\times 2$$

Multiply 9 and 2 to get 18.

$$16+a^{2}=18$$

Subtract 16 from both sides.

$$a^{2}=18-16$$

Subtract 16 from 18 to get 2.

$$a^{2}=2$$

Take the square root of both sides of the equation.

$$a=\sqrt{2} a=-\sqrt{2}$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$16+a^{2}=\left(3\sqrt{2}\right)^{2}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$16+a^{2}=3^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$16+a^{2}=9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$16+a^{2}=9\times 2$$

Multiply 9 and 2 to get 18.

$$16+a^{2}=18$$

Subtract 18 from both sides.

$$16+a^{2}-18=0$$

Subtract 18 from 16 to get -2.

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

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.

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

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

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

Square 0.

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

Multiply -4 times -2.

$$a=\frac{0±\sqrt{8}}{2}$$

Take the square root of 8.

$$a=\frac{0±2\sqrt{2}}{2}$$

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

$$a=\sqrt{2}$$

Now solve the equation a=\frac{0±2\sqrt{2}}{2} when ± is minus.

$$a=-\sqrt{2}$$

The equation is now solved.

$$a=\sqrt{2} a=-\sqrt{2}$$