How to solve 5(x)^(2)=3x

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 5(x)^(2)=3x | Equatix

Question

$$5 { x }^{ 2 } =3x$$

Solve for x

$x=\frac{3}{5}=0.6$
$x=0$

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Subtract $3x$ from both sides.

$$5x^{2}-3x=0$$

Factor out $x$.

$$x\left(5x-3\right)=0$$

To find equation solutions, solve $x=0$ and $5x-3=0$.

$$x=0$$ $$x=\frac{3}{5}$$

Steps Using the Quadratic Formula

Subtract $3x$ from both sides.

$$5x^{2}-3x=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $-3$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}}}{2\times 5}$$

Take the square root of $\left(-3\right)^{2}$.

$$x=\frac{-\left(-3\right)±3}{2\times 5}$$

The opposite of $-3$ is $3$.

$$x=\frac{3±3}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{3±3}{10}$$

Now solve the equation $x=\frac{3±3}{10}$ when $±$ is plus. Add $3$ to $3$.

$$x=\frac{6}{10}$$

Reduce the fraction $\frac{6}{10}$ to lowest terms by extracting and canceling out $2$.

$$x=\frac{3}{5}$$

Now solve the equation $x=\frac{3±3}{10}$ when $±$ is minus. Subtract $3$ from $3$.

$$x=\frac{0}{10}$$

Divide $0$ by $10$.

$$x=0$$

The equation is now solved.

$$x=\frac{3}{5}$$ $$x=0$$

Steps for Completing the Square

Subtract $3x$ from both sides.

$$5x^{2}-3x=0$$

Divide both sides by $5$.

$$\frac{5x^{2}-3x}{5}=\frac{0}{5}$$

Dividing by $5$ undoes the multiplication by $5$.

$$x^{2}-\frac{3}{5}x=\frac{0}{5}$$

Divide $0$ by $5$.

$$x^{2}-\frac{3}{5}x=0$$

Divide $-\frac{3}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{3}{10}$. Then add the square of $-\frac{3}{10}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{5}x+\left(-\frac{3}{10}\right)^{2}=\left(-\frac{3}{10}\right)^{2}$$

Square $-\frac{3}{10}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{5}x+\frac{9}{100}=\frac{9}{100}$$

Factor $x^{2}-\frac{3}{5}x+\frac{9}{100}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{3}{10}\right)^{2}=\frac{9}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{10}\right)^{2}}=\sqrt{\frac{9}{100}}$$

Simplify.

$$x-\frac{3}{10}=\frac{3}{10}$$ $$x-\frac{3}{10}=-\frac{3}{10}$$

Add $\frac{3}{10}$ to both sides of the equation.

$$x=\frac{3}{5}$$ $$x=0$$

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Question

Solve for x

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Steps Using the Quadratic Formula

Steps for Completing the Square