Solve for a (x+a)^2=x^2+16x+64

${(x + a)}^{2} = x^{2} + 16 x + 64$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x + a = \pm \sqrt{x^{2} + 16 x + 64}$

Simplify $\pm \sqrt{x^{2} + 16 x + 64}$ .

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$x + a = \pm \sqrt{x^{2} + 16 x + 8^{2}}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

Rewrite the polynomial.

$x + a = \pm \sqrt{x^{2} + 2 \cdot x \cdot 8 + 8^{2}}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 8$ .

$x + a = \pm \sqrt{{(x + 8)}^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x + a = \pm (x + 8)$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x + a = x + 8$

Move all terms not containing $a$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

$a = x + 8 - x$

Combine the opposite terms in $x + 8 - x$ .

Subtract $x$ from $x$ .

$a = 0 + 8$

Add $0$ and $8$ .

$a = 8$

Next, use the negative value of the $\pm$ to find the second solution.

$x + a = - (x + 8)$

Simplify $- (x + 8)$ .

Rewrite.

$x + a = 0 + 0 - (x + 8)$

Simplify by adding zeros.

$x + a = - (x + 8)$

Apply the distributive property.

$x + a = - x - 1 \cdot 8$

Multiply $- 1$ by $8$ .

$x + a = - x - 8$

Move all terms not containing $a$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

$a = - x - 8 - x$

Subtract $x$ from $- x$ .

$a = - 2 x - 8$

The complete solution is the result of both the positive and negative portions of the solution.

$a = 8$

$a = - 2 x - 8$

$a = 8$

$a = - 2 x - 8$

Solve for a (x+a)^2=x^2+16x+64

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