Solve (sqrt{x+16})^2=(x+1)^2

$$( \sqrt { x + 16 } ) ^ { 2 } = ( x + 1 ) ^ { 2 }$$

Solve for x

$$x = \frac{\sqrt{61} - 1}{2} \approx 3.405124838$$

$$x=\frac{-\sqrt{61}-1}{2}\approx -4.405124838$$

Steps Using the Quadratic Formula

Calculate \sqrt{x+16} to the power of 2 and get x+16.

$$x+16=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x+16=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$x+16-x^{2}=2x+1$$

Subtract 2x from both sides.

$$x+16-x^{2}-2x=1$$

Combine x and -2x to get -x.

$$-x+16-x^{2}=1$$

Subtract 1 from both sides.

$$-x+16-x^{2}-1=0$$

Subtract 1 from 16 to get 15.

$$-x+15-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -1 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-1\right)\times 15}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-1\right)±\sqrt{1+4\times 15}}{2\left(-1\right)}$$

Multiply 4 times 15.

$$x=\frac{-\left(-1\right)±\sqrt{1+60}}{2\left(-1\right)}$$

Add 1 to 60.

$$x=\frac{-\left(-1\right)±\sqrt{61}}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{61}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±\sqrt{61}}{-2}$$

Now solve the equation x=\frac{1±\sqrt{61}}{-2} when ± is plus. Add 1 to \sqrt{61}.

$$x=\frac{\sqrt{61}+1}{-2}$$

Divide 1+\sqrt{61} by -2.

$$x=\frac{-\sqrt{61}-1}{2}$$

Now solve the equation x=\frac{1±\sqrt{61}}{-2} when ± is minus. Subtract \sqrt{61} from 1.

$$x=\frac{1-\sqrt{61}}{-2}$$

Divide 1-\sqrt{61} by -2.

$$x=\frac{\sqrt{61}-1}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{61}-1}{2} x=\frac{\sqrt{61}-1}{2}$$

Steps for Completing the Square

Calculate \sqrt{x+16} to the power of 2 and get x+16.

$$x+16=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x+16=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$x+16-x^{2}=2x+1$$

Subtract 2x from both sides.

$$x+16-x^{2}-2x=1$$

Combine x and -2x to get -x.

$$-x+16-x^{2}=1$$

Subtract 16 from both sides.

$$-x-x^{2}=1-16$$

Subtract 16 from 1 to get -15.

$$-x-x^{2}=-15$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-x=-15$$

Divide both sides by -1.

$$\frac{-x^{2}-x}{-1}=-\frac{15}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{1}{-1}\right)x=-\frac{15}{-1}$$

Divide -1 by -1.

$$x^{2}+x=-\frac{15}{-1}$$

Divide -15 by -1.

$$x^{2}+x=15$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=15+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=15+\frac{1}{4}$$

Add 15 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{61}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{61}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{61}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{61}}{2} x+\frac{1}{2}=-\frac{\sqrt{61}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{61}-1}{2} x=\frac{-\sqrt{61}-1}{2}$$