Solve Date x ^ 2 - 2x ^ 2 + 4x - 8 = 0 2 | 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

$$x ^ { 2 } - 2 x ^ { 2 } + 4 x - 8 = 0$$

Solve for x (complex solution)

$x=2-2i$
$x=2+2i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

$$-x^{2}+4x-8=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-1$ for $a$, $4$ for $b$, and $-8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Square $4$.

$$x=\frac{-4±\sqrt{16-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}$$

Multiply $-4$ times $-1$.

$$x=\frac{-4±\sqrt{16+4\left(-8\right)}}{2\left(-1\right)}$$

Multiply $4$ times $-8$.

$$x=\frac{-4±\sqrt{16-32}}{2\left(-1\right)}$$

Add $16$ to $-32$.

$$x=\frac{-4±\sqrt{-16}}{2\left(-1\right)}$$

Take the square root of $-16$.

$$x=\frac{-4±4i}{2\left(-1\right)}$$

Multiply $2$ times $-1$.

$$x=\frac{-4±4i}{-2}$$

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

$$x=\frac{-4+4i}{-2}$$

Divide $-4+4i$ by $-2$.

$$x=2-2i$$

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

$$x=\frac{-4-4i}{-2}$$

Divide $-4-4i$ by $-2$.

$$x=2+2i$$

The equation is now solved.

$$x=2-2i$$ $$x=2+2i$$

Steps for Completing the Square

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

$$-x^{2}+4x-8=0$$

Add $8$ to both sides. Anything plus zero gives itself.

$$-x^{2}+4x=8$$

Divide both sides by $-1$.

$$\frac{-x^{2}+4x}{-1}=\frac{8}{-1}$$

Dividing by $-1$ undoes the multiplication by $-1$.

$$x^{2}+\frac{4}{-1}x=\frac{8}{-1}$$

Divide $4$ by $-1$.

$$x^{2}-4x=\frac{8}{-1}$$

Divide $8$ by $-1$.

$$x^{2}-4x=-8$$

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

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

Square $-2$.

$$x^{2}-4x+4=-8+4$$

Add $-8$ to $4$.

$$x^{2}-4x+4=-4$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=2i$$ $$x-2=-2i$$

Add $2$ to both sides of the equation.

$$x=2+2i$$ $$x=2-2i$$