How to solve 6x - 1 = 2x ^ 2

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 6x - 1 = 2x ^ 2 | Equatix

Question

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

Solve for x

$x = \frac{\sqrt{7} + 3}{2} \approx 2.822875656$
$x=\frac{3-\sqrt{7}}{2}\approx 0.177124344$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

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

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

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.

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

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

$$x=\frac{-6±\sqrt{6^{2}-4\left(-2\right)\left(-1\right)}}{2\left(-2\right)}$$

Square $6$.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)\left(-1\right)}}{2\left(-2\right)}$$

Multiply $-4$ times $-2$.

$$x=\frac{-6±\sqrt{36+8\left(-1\right)}}{2\left(-2\right)}$$

Multiply $8$ times $-1$.

$$x=\frac{-6±\sqrt{36-8}}{2\left(-2\right)}$$

Add $36$ to $-8$.

$$x=\frac{-6±\sqrt{28}}{2\left(-2\right)}$$

Take the square root of $28$.

$$x=\frac{-6±2\sqrt{7}}{2\left(-2\right)}$$

Multiply $2$ times $-2$.

$$x=\frac{-6±2\sqrt{7}}{-4}$$

Now solve the equation $x=\frac{-6±2\sqrt{7}}{-4}$ when $±$ is plus. Add $-6$ to $2\sqrt{7}$.

$$x=\frac{2\sqrt{7}-6}{-4}$$

Divide $-6+2\sqrt{7}$ by $-4$.

$$x=\frac{3-\sqrt{7}}{2}$$

Now solve the equation $x=\frac{-6±2\sqrt{7}}{-4}$ when $±$ is minus. Subtract $2\sqrt{7}$ from $-6$.

$$x=\frac{-2\sqrt{7}-6}{-4}$$

Divide $-6-2\sqrt{7}$ by $-4$.

$$x=\frac{\sqrt{7}+3}{2}$$

The equation is now solved.

$$x=\frac{3-\sqrt{7}}{2}$$ $$x=\frac{\sqrt{7}+3}{2}$$

Steps for Completing the Square

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

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

Add $1$ to both sides. Anything plus zero gives itself.

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

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

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

Divide both sides by $-2$.

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

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

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

Divide $6$ by $-2$.

$$x^{2}-3x=\frac{1}{-2}$$

Divide $1$ by $-2$.

$$x^{2}-3x=-\frac{1}{2}$$

Divide $-3$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{2}$. Then add the square of $-\frac{3}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square $-\frac{3}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-\frac{1}{2}+\frac{9}{4}$$

Add $-\frac{1}{2}$ to $\frac{9}{4}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{7}{4}$$

Factor $x^{2}-3x+\frac{9}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{7}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{7}}{2}$$ $$x-\frac{3}{2}=-\frac{\sqrt{7}}{2}$$

Add $\frac{3}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{7}+3}{2}$$ $$x=\frac{3-\sqrt{7}}{2}$$

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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