How to solve x/2 = x ^ 2 - 1

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/2 = x ^ 2 - 1 | Equatix

Question

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

Solve for x

$x = \frac{\sqrt{17} + 1}{4} \approx 1.280776406$
$x=\frac{1-\sqrt{17}}{4}\approx -0.780776406$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Multiply both sides of the equation by $2$.

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

Subtract $2x^{2}$ from both sides.

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

Add $2$ to both sides.

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

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.

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

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

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

Square $1$.

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

Multiply $-4$ times $-2$.

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

Multiply $8$ times $2$.

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

Add $1$ to $16$.

$$x=\frac{-1±\sqrt{17}}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{-1±\sqrt{17}}{-4}$$

Now solve the equation $x=\frac{-1±\sqrt{17}}{-4}$ when $±$ is plus. Add $-1$ to $\sqrt{17}$.

$$x=\frac{\sqrt{17}-1}{-4}$$

Divide $-1+\sqrt{17}$ by $-4$.

$$x=\frac{1-\sqrt{17}}{4}$$

Now solve the equation $x=\frac{-1±\sqrt{17}}{-4}$ when $±$ is minus. Subtract $\sqrt{17}$ from $-1$.

$$x=\frac{-\sqrt{17}-1}{-4}$$

Divide $-1-\sqrt{17}$ by $-4$.

$$x=\frac{\sqrt{17}+1}{4}$$

The equation is now solved.

$$x=\frac{1-\sqrt{17}}{4}$$ $$x=\frac{\sqrt{17}+1}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by $2$.

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

Subtract $2x^{2}$ from both sides.

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

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

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

Divide both sides by $-2$.

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

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

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

Divide $1$ by $-2$.

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

Divide $-2$ by $-2$.

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

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.

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

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

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

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{17}{16}$$

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 $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{17}}{4}$$ $$x-\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

Add $\frac{1}{4}$ to both sides of the equation.

$$x=\frac{\sqrt{17}+1}{4}$$ $$x=\frac{1-\sqrt{17}}{4}$$

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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