Solve x^2+(x+1)^2=31

$$x ^ { 2 } + ( x + 1 ) ^ { 2 } = 31$$

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

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

$$x^{2}+x^{2}+2x+1=31$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+2x+1=31$$

Subtract 31 from both sides.

$$2x^{2}+2x+1-31=0$$

Subtract 31 from 1 to get -30.

$$2x^{2}+2x-30=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -30 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-30\right)}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-30\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-30\right)}}{2\times 2}$$

Multiply -8 times -30.

$$x=\frac{-2±\sqrt{4+240}}{2\times 2}$$

Add 4 to 240.

$$x=\frac{-2±\sqrt{244}}{2\times 2}$$

Take the square root of 244.

$$x=\frac{-2±2\sqrt{61}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±2\sqrt{61}}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{61}}{4} when ± is plus. Add -2 to 2\sqrt{61}.

$$x=\frac{2\sqrt{61}-2}{4}$$

Divide -2+2\sqrt{61} by 4.

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

Now solve the equation x=\frac{-2±2\sqrt{61}}{4} when ± is minus. Subtract 2\sqrt{61} from -2.

$$x=\frac{-2\sqrt{61}-2}{4}$$

Divide -2-2\sqrt{61} by 4.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{2}+x^{2}+2x+1=31$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+2x+1=31$$

Subtract 1 from both sides.

$$2x^{2}+2x=31-1$$

Subtract 1 from 31 to get 30.

$$2x^{2}+2x=30$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{30}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{30}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{30}{2}$$

Divide 30 by 2.

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

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.

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