Solve x ^ 2 + (x + 1) ^ 2 = 244 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x^{2}+(x+1)^{2}=244$$

Solve for x

$x = \frac{\sqrt{487} - 1}{2} \approx 10.534038245$
$x=\frac{-\sqrt{487}-1}{2}\approx -11.534038245$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$x^{2}+x^{2}+2x+1=244$$

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

$$2x^{2}+2x+1=244$$

Subtract $244$ from both sides.

$$2x^{2}+2x+1-244=0$$

Subtract $244$ from $1$ to get $-243$.

$$2x^{2}+2x-243=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $2$ for $b$, and $-243$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-243\right)}}{2\times 2}$$

Square $2$.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-243\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-2±\sqrt{4-8\left(-243\right)}}{2\times 2}$$

Multiply $-8$ times $-243$.

$$x=\frac{-2±\sqrt{4+1944}}{2\times 2}$$

Add $4$ to $1944$.

$$x=\frac{-2±\sqrt{1948}}{2\times 2}$$

Take the square root of $1948$.

$$x=\frac{-2±2\sqrt{487}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{-2±2\sqrt{487}}{4}$$

Now solve the equation $x=\frac{-2±2\sqrt{487}}{4}$ when $±$ is plus. Add $-2$ to $2\sqrt{487}$.

$$x=\frac{2\sqrt{487}-2}{4}$$

Divide $-2+2\sqrt{487}$ by $4$.

$$x=\frac{\sqrt{487}-1}{2}$$

Now solve the equation $x=\frac{-2±2\sqrt{487}}{4}$ when $±$ is minus. Subtract $2\sqrt{487}$ from $-2$.

$$x=\frac{-2\sqrt{487}-2}{4}$$

Divide $-2-2\sqrt{487}$ by $4$.

$$x=\frac{-\sqrt{487}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{487}-1}{2}$$ $$x=\frac{-\sqrt{487}-1}{2}$$

Steps for Completing the Square

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

$$x^{2}+x^{2}+2x+1=244$$

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

$$2x^{2}+2x+1=244$$

Subtract $1$ from both sides.

$$2x^{2}+2x=244-1$$

Subtract $1$ from $244$ to get $243$.

$$2x^{2}+2x=243$$

Divide both sides by $2$.

$$\frac{2x^{2}+2x}{2}=\frac{243}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$x^{2}+\frac{2}{2}x=\frac{243}{2}$$

Divide $2$ by $2$.

$$x^{2}+x=\frac{243}{2}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{243}{2}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{243}{2}+\frac{1}{4}$$

Add $\frac{243}{2}$ to $\frac{1}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{487}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{487}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{487}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{487}}{2}$$ $$x+\frac{1}{2}=-\frac{\sqrt{487}}{2}$$

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

$$x=\frac{\sqrt{487}-1}{2}$$ $$x=\frac{-\sqrt{487}-1}{2}$$