Solve x^2+(x+1)^2=4191

$$x ^ { 2 } + ( x + 1 ) ^ { 2 } = 4191$$

Solve for x

$$x = \frac{17 \sqrt{29} - 1}{2} \approx 45.273900861$$

$$x=\frac{-17\sqrt{29}-1}{2}\approx -46.273900861$$

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}.

$$x^{2}+x^{2}+2x+1=4191$$

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

$$2x^{2}+2x+1=4191$$

Subtract 4191 from both sides.

$$2x^{2}+2x+1-4191=0$$

Subtract 4191 from 1 to get -4190.

$$2x^{2}+2x-4190=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -4190 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-4190\right)}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-4190\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-4190\right)}}{2\times 2}$$

Multiply -8 times -4190.

$$x=\frac{-2±\sqrt{4+33520}}{2\times 2}$$

Add 4 to 33520.

$$x=\frac{-2±\sqrt{33524}}{2\times 2}$$

Take the square root of 33524.

$$x=\frac{-2±34\sqrt{29}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±34\sqrt{29}}{4}$$

Now solve the equation x=\frac{-2±34\sqrt{29}}{4} when ± is plus. Add -2 to 34\sqrt{29}.

$$x=\frac{34\sqrt{29}-2}{4}$$

Divide -2+34\sqrt{29} by 4.

$$x=\frac{17\sqrt{29}-1}{2}$$

Now solve the equation x=\frac{-2±34\sqrt{29}}{4} when ± is minus. Subtract 34\sqrt{29} from -2.

$$x=\frac{-34\sqrt{29}-2}{4}$$

Divide -2-34\sqrt{29} by 4.

$$x=\frac{-17\sqrt{29}-1}{2}$$

The equation is now solved.

$$x=\frac{17\sqrt{29}-1}{2} x=\frac{-17\sqrt{29}-1}{2}$$

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}.

$$x^{2}+x^{2}+2x+1=4191$$

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

$$2x^{2}+2x+1=4191$$

Subtract 1 from both sides.

$$2x^{2}+2x=4191-1$$

Subtract 1 from 4191 to get 4190.

$$2x^{2}+2x=4190$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{4190}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{4190}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{4190}{2}$$

Divide 4190 by 2.

$$x^{2}+x=2095$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=2095+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=2095+\frac{1}{4}$$

Add 2095 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{8381}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{8381}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{8381}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{17\sqrt{29}}{2} x+\frac{1}{2}=-\frac{17\sqrt{29}}{2}$$

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

$$x=\frac{17\sqrt{29}-1}{2} x=\frac{-17\sqrt{29}-1}{2}$$