Solve (x + 1) ^ 2 = 2(x - 3) | 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

$$(x+1)^{2}=2(x-3)$$

Solve for x (complex solution)

$x=-\sqrt{7}i\approx -0-2.645751311i$
$x=\sqrt{7}i\approx 2.645751311i$

Steps by Finding Square Root

Steps Using the Quadratic Formula

Steps by Finding Square Root

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

$$x^{2}+2x+1=2\left(x-3\right)$$

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

$$x^{2}+2x+1=2x-6$$

Subtract $2x$ from both sides.

$$x^{2}+2x+1-2x=-6$$

Combine $2x$ and $-2x$ to get $0$.

$$x^{2}+1=-6$$

Subtract $1$ from both sides.

$$x^{2}=-6-1$$

Subtract $1$ from $-6$ to get $-7$.

$$x^{2}=-7$$

The equation is now solved.

$$x=\sqrt{7}i$$ $$x=-\sqrt{7}i$$

Steps Using the Quadratic Formula

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

$$x^{2}+2x+1=2\left(x-3\right)$$

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

$$x^{2}+2x+1=2x-6$$

Subtract $2x$ from both sides.

$$x^{2}+2x+1-2x=-6$$

Combine $2x$ and $-2x$ to get $0$.

$$x^{2}+1=-6$$

Add $6$ to both sides.

$$x^{2}+1+6=0$$

Add $1$ and $6$ to get $7$.

$$x^{2}+7=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $7$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{0±\sqrt{0^{2}-4\times 7}}{2}$$

Square $0$.

$$x=\frac{0±\sqrt{-4\times 7}}{2}$$

Multiply $-4$ times $7$.

$$x=\frac{0±\sqrt{-28}}{2}$$

Take the square root of $-28$.

$$x=\frac{0±2\sqrt{7}i}{2}$$

Now solve the equation $x=\frac{0±2\sqrt{7}i}{2}$ when $±$ is plus.

$$x=\sqrt{7}i$$

Now solve the equation $x=\frac{0±2\sqrt{7}i}{2}$ when $±$ is minus.

$$x=-\sqrt{7}i$$

The equation is now solved.

$$x=\sqrt{7}i$$ $$x=-\sqrt{7}i$$