Solve Using the Quadratic Formula a^2+(4a)^2=15^2

$a^{2} + {(4 a)}^{2} = 15^{2}$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $a^{2} + {(4 a)}^{2}$ .

Simplify each term.

Apply the product rule to $4 a$ .

$a^{2} + 4^{2} a^{2} = 15^{2}$

Raise $4$ to the power of $2$ .

$a^{2} + 16 a^{2} = 15^{2}$

Add $a^{2}$ and $16 a^{2}$ .

$17 a^{2} = 15^{2}$

Simplify the right side.

Raise $15$ to the power of $2$ .

$17 a^{2} = 225$

Subtract $225$ from both sides of the equation.

$17 a^{2} - 225 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 17$ , $b = 0$ , and $c = - 225$ into the quadratic formula and solve for $a$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (17 \cdot - 225)}}{2 \cdot 17}$

Step 4

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$a = \frac{0 \pm \sqrt{0 - 4 \cdot 17 \cdot - 225}}{2 \cdot 17}$

Multiply $- 4 \cdot 17 \cdot - 225$ .

Multiply $- 4$ by $17$ .

$a = \frac{0 \pm \sqrt{0 - 68 \cdot - 225}}{2 \cdot 17}$

Multiply $- 68$ by $- 225$ .

$a = \frac{0 \pm \sqrt{0 + 15300}}{2 \cdot 17}$

Add $0$ and $15300$ .

$a = \frac{0 \pm \sqrt{15300}}{2 \cdot 17}$

Rewrite $15300$ as $30^{2} \cdot 17$ .

Factor $900$ out of $15300$ .

$a = \frac{0 \pm \sqrt{900 (17)}}{2 \cdot 17}$

Rewrite $900$ as $30^{2}$ .

$a = \frac{0 \pm \sqrt{30^{2} \cdot 17}}{2 \cdot 17}$

Pull terms out from under the radical.

$a = \frac{0 \pm 30 \sqrt{17}}{2 \cdot 17}$

Multiply $2$ by $17$ .

$a = \frac{0 \pm 30 \sqrt{17}}{34}$

Simplify $\frac{0 \pm 30 \sqrt{17}}{34}$ .

$a = \frac{\pm 15 \sqrt{17}}{17}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$a = \frac{\pm 15 \sqrt{17}}{17}$

Decimal Form:

$a = 3.63803437 \dots, - 3.63803437 \dots$

Solve Using the Quadratic Formula a^2+(4a)^2=15^2

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