Solve ( ii ) x ^ 2 - 2x = (- 2)(3 - x) | 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

$$x ^ { 2 } - 2 x = ( - 2 ) ( 3 - x )$$

Solve for x (complex solution)

$x=2+\sqrt{2}i\approx 2+1.414213562i$
$x=-\sqrt{2}i+2\approx 2-1.414213562i$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Use the distributive property to multiply $-2$ by $3-x$.

$$x^{2}-2x=-6+2x$$

Subtract $-6$ from both sides.

$$x^{2}-2x-\left(-6\right)=2x$$

The opposite of $-6$ is $6$.

$$x^{2}-2x+6=2x$$

Subtract $2x$ from both sides.

$$x^{2}-2x+6-2x=0$$

Combine $-2x$ and $-2x$ to get $-4x$.

$$x^{2}-4x+6=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-4$ for $b$, and $6$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 6}}{2}$$

Square $-4$.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 6}}{2}$$

Multiply $-4$ times $6$.

$$x=\frac{-\left(-4\right)±\sqrt{16-24}}{2}$$

Add $16$ to $-24$.

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

Take the square root of $-8$.

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

The opposite of $-4$ is $4$.

$$x=\frac{4±2\sqrt{2}i}{2}$$

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

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

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

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

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply $-2$ by $3-x$.

$$x^{2}-2x=-6+2x$$

Subtract $2x$ from both sides.

$$x^{2}-2x-2x=-6$$

Combine $-2x$ and $-2x$ to get $-4x$.

$$x^{2}-4x=-6$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-6+\left(-2\right)^{2}$$

Square $-2$.

$$x^{2}-4x+4=-6+4$$

Add $-6$ to $4$.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add $2$ to both sides of the equation.

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