How to solve (x-1)^(2)=2x+3

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

Question

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

Solve for x

$x=\sqrt{6}+2\approx 4.449489743$
$x=2-\sqrt{6}\approx -0.449489743$

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 $\left(x-1\right)^{2}$.

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

Subtract $2x$ from both sides.

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

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

$$x^{2}-4x+1=3$$

Subtract $3$ from both sides.

$$x^{2}-4x+1-3=0$$

Subtract $3$ from $1$ to get $-2$.

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

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

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

Square $-4$.

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

Multiply $-4$ times $-2$.

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

Add $16$ to $8$.

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

Take the square root of $24$.

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

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

$$x=\frac{4±2\sqrt{6}}{2}$$

Now solve the equation $x=\frac{4±2\sqrt{6}}{2}$ when $±$ is plus. Add $4$ to $2\sqrt{6}$.

$$x=\frac{2\sqrt{6}+4}{2}$$

Divide $4+2\sqrt{6}$ by $2$.

$$x=\sqrt{6}+2$$

Now solve the equation $x=\frac{4±2\sqrt{6}}{2}$ when $±$ is minus. Subtract $2\sqrt{6}$ from $4$.

$$x=\frac{4-2\sqrt{6}}{2}$$

Divide $4-2\sqrt{6}$ by $2$.

$$x=2-\sqrt{6}$$

The equation is now solved.

$$x=\sqrt{6}+2$$ $$x=2-\sqrt{6}$$

Steps for Completing the Square

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

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

Subtract $2x$ from both sides.

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

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

$$x^{2}-4x+1=3$$

Subtract $1$ from both sides.

$$x^{2}-4x=3-1$$

Subtract $1$ from $3$ to get $2$.

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

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.

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

Square $-2$.

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

Add $2$ to $4$.

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

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=\sqrt{6}$$ $$x-2=-\sqrt{6}$$

Add $2$ to both sides of the equation.

$$x=\sqrt{6}+2$$ $$x=2-\sqrt{6}$$

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Example

Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square