How to solve 2 x = 5x + 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 2 x = 5x + 2 | Equatix

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Question

$$x ^ { 2 } = 5 x + 2$$

Solve for x

$x = \frac{\sqrt{33} + 5}{2} \approx 5.372281323$
$x=\frac{5-\sqrt{33}}{2}\approx -0.372281323$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Subtract $5x$ from both sides.

$$x^{2}-5x=2$$

Subtract $2$ from both sides.

$$x^{2}-5x-2=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-5$ for $b$, and $-2$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)}}{2}$$

Square $-5$.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)}}{2}$$

Multiply $-4$ times $-2$.

$$x=\frac{-\left(-5\right)±\sqrt{25+8}}{2}$$

Add $25$ to $8$.

$$x=\frac{-\left(-5\right)±\sqrt{33}}{2}$$

The opposite of $-5$ is $5$.

$$x=\frac{5±\sqrt{33}}{2}$$

Now solve the equation $x=\frac{5±\sqrt{33}}{2}$ when $±$ is plus. Add $5$ to $\sqrt{33}$.

$$x=\frac{\sqrt{33}+5}{2}$$

Now solve the equation $x=\frac{5±\sqrt{33}}{2}$ when $±$ is minus. Subtract $\sqrt{33}$ from $5$.

$$x=\frac{5-\sqrt{33}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+5}{2}$$ $$x=\frac{5-\sqrt{33}}{2}$$

Steps for Completing the Square

Subtract $5x$ from both sides.

$$x^{2}-5x=2$$

Divide $-5$, the coefficient of the $x$ term, by $2$ to get $-\frac{5}{2}$. Then add the square of $-\frac{5}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=2+\left(-\frac{5}{2}\right)^{2}$$

Square $-\frac{5}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=2+\frac{25}{4}$$

Add $2$ to $\frac{25}{4}$.

$$x^{2}-5x+\frac{25}{4}=\frac{33}{4}$$

Factor $x^{2}-5x+\frac{25}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{33}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{33}}{2}$$ $$x-\frac{5}{2}=-\frac{\sqrt{33}}{2}$$

Add $\frac{5}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{33}+5}{2}$$ $$x=\frac{5-\sqrt{33}}{2}$$

Show Solution

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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