Solve (x)^(2)+(x-8)^(2)=36 | 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

$${ x }^{ 2 } + { \left(x-8 \right) }^{ 2 } =36$$

Solve for x

$x=\sqrt{2}+4\approx 5.414213562$
$x=4-\sqrt{2}\approx 2.585786438$

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 $\left(x-8\right)^{2}$.

$$x^{2}+x^{2}-16x+64=36$$

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

$$2x^{2}-16x+64=36$$

Subtract $36$ from both sides.

$$2x^{2}-16x+64-36=0$$

Subtract $36$ from $64$ to get $28$.

$$2x^{2}-16x+28=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-16$ for $b$, and $28$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 28}}{2\times 2}$$

Square $-16$.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 28}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-\left(-16\right)±\sqrt{256-8\times 28}}{2\times 2}$$

Multiply $-8$ times $28$.

$$x=\frac{-\left(-16\right)±\sqrt{256-224}}{2\times 2}$$

Add $256$ to $-224$.

$$x=\frac{-\left(-16\right)±\sqrt{32}}{2\times 2}$$

Take the square root of $32$.

$$x=\frac{-\left(-16\right)±4\sqrt{2}}{2\times 2}$$

The opposite of $-16$ is $16$.

$$x=\frac{16±4\sqrt{2}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{16±4\sqrt{2}}{4}$$

Now solve the equation $x=\frac{16±4\sqrt{2}}{4}$ when $±$ is plus. Add $16$ to $4\sqrt{2}$.

$$x=\frac{4\sqrt{2}+16}{4}$$

Divide $16+4\sqrt{2}$ by $4$.

$$x=\sqrt{2}+4$$

Now solve the equation $x=\frac{16±4\sqrt{2}}{4}$ when $±$ is minus. Subtract $4\sqrt{2}$ from $16$.

$$x=\frac{16-4\sqrt{2}}{4}$$

Divide $16-4\sqrt{2}$ by $4$.

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

The equation is now solved.

$$x=\sqrt{2}+4$$ $$x=4-\sqrt{2}$$

Steps for Completing the Square

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-8\right)^{2}$.

$$x^{2}+x^{2}-16x+64=36$$

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

$$2x^{2}-16x+64=36$$

Subtract $64$ from both sides.

$$2x^{2}-16x=36-64$$

Subtract $64$ from $36$ to get $-28$.

$$2x^{2}-16x=-28$$

Divide both sides by $2$.

$$\frac{2x^{2}-16x}{2}=-\frac{28}{2}$$

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

$$x^{2}+\left(-\frac{16}{2}\right)x=-\frac{28}{2}$$

Divide $-16$ by $2$.

$$x^{2}-8x=-\frac{28}{2}$$

Divide $-28$ by $2$.

$$x^{2}-8x=-14$$

Divide $-8$, the coefficient of the $x$ term, by $2$ to get $-4$. Then add the square of $-4$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=-14+\left(-4\right)^{2}$$

Square $-4$.

$$x^{2}-8x+16=-14+16$$

Add $-14$ to $16$.

$$x^{2}-8x+16=2$$

Factor $x^{2}-8x+16$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-4\right)^{2}=2$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{2}$$

Simplify.

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

Add $4$ to both sides of the equation.

$$x=\sqrt{2}+4$$ $$x=4-\sqrt{2}$$