How to solve x + 2xx + 2 = 256

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve x + 2xx + 2 = 256 | Equatix

Question

$$x+2xx+2=256$$

Solve for x

$x = \frac{\sqrt{2033} - 1}{4} \approx 11.022200318$
$x=\frac{-\sqrt{2033}-1}{4}\approx -11.522200318$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$x+2x^{2}+2=256$$

Subtract $256$ from both sides.

$$x+2x^{2}+2-256=0$$

Subtract $256$ from $2$ to get $-254$.

$$x+2x^{2}-254=0$$

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.

$$2x^{2}+x-254=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $1$ for $b$, and $-254$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-254\right)}}{2\times 2}$$

Square $1$.

$$x=\frac{-1±\sqrt{1-4\times 2\left(-254\right)}}{2\times 2}$$

Multiply $-4$ times $2$.

$$x=\frac{-1±\sqrt{1-8\left(-254\right)}}{2\times 2}$$

Multiply $-8$ times $-254$.

$$x=\frac{-1±\sqrt{1+2032}}{2\times 2}$$

Add $1$ to $2032$.

$$x=\frac{-1±\sqrt{2033}}{2\times 2}$$

Multiply $2$ times $2$.

$$x=\frac{-1±\sqrt{2033}}{4}$$

Now solve the equation $x=\frac{-1±\sqrt{2033}}{4}$ when $±$ is plus. Add $-1$ to $\sqrt{2033}$.

$$x=\frac{\sqrt{2033}-1}{4}$$

Now solve the equation $x=\frac{-1±\sqrt{2033}}{4}$ when $±$ is minus. Subtract $\sqrt{2033}$ from $-1$.

$$x=\frac{-\sqrt{2033}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{2033}-1}{4}$$ $$x=\frac{-\sqrt{2033}-1}{4}$$

Steps for Completing the Square

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

$$x+2x^{2}+2=256$$

Subtract $2$ from both sides.

$$x+2x^{2}=256-2$$

Subtract $2$ from $256$ to get $254$.

$$x+2x^{2}=254$$

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 $x^{2}+bx=c$.

$$2x^{2}+x=254$$

Divide both sides by $2$.

$$\frac{2x^{2}+x}{2}=\frac{254}{2}$$

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

$$x^{2}+\frac{1}{2}x=\frac{254}{2}$$

Divide $254$ by $2$.

$$x^{2}+\frac{1}{2}x=127$$

Divide $\frac{1}{2}$, the coefficient of the $x$ term, by $2$ to get $\frac{1}{4}$. Then add the square of $\frac{1}{4}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=127+\left(\frac{1}{4}\right)^{2}$$

Square $\frac{1}{4}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=127+\frac{1}{16}$$

Add $127$ to $\frac{1}{16}$.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{2033}{16}$$

Factor $x^{2}+\frac{1}{2}x+\frac{1}{16}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{2033}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{2033}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{2033}}{4}$$ $$x+\frac{1}{4}=-\frac{\sqrt{2033}}{4}$$

Subtract $\frac{1}{4}$ from both sides of the equation.

$$x=\frac{\sqrt{2033}-1}{4}$$ $$x=\frac{-\sqrt{2033}-1}{4}$$

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Steps Using the Quadratic Formula

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