Solve (84(x + 5 - x))/(x(x + 5)) = 5 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{84(x+5-x)}{x(x+5)}=5$$

Solve for x

$x=-12$
$x=7$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to any of the values $-5,0$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x+5\right)$.

$$84\left(x+5-x\right)=5x\left(x+5\right)$$

Combine $x$ and $-x$ to get $0$.

$$84\times 5=5x\left(x+5\right)$$

Multiply $84$ and $5$ to get $420$.

$$420=5x\left(x+5\right)$$

Use the distributive property to multiply $5x$ by $x+5$.

$$420=5x^{2}+25x$$

Swap sides so that all variable terms are on the left hand side.

$$5x^{2}+25x=420$$

Subtract $420$ from both sides.

$$5x^{2}+25x-420=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $25$ for $b$, and $-420$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-25±\sqrt{25^{2}-4\times 5\left(-420\right)}}{2\times 5}$$

Square $25$.

$$x=\frac{-25±\sqrt{625-4\times 5\left(-420\right)}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-25±\sqrt{625-20\left(-420\right)}}{2\times 5}$$

Multiply $-20$ times $-420$.

$$x=\frac{-25±\sqrt{625+8400}}{2\times 5}$$

Add $625$ to $8400$.

$$x=\frac{-25±\sqrt{9025}}{2\times 5}$$

Take the square root of $9025$.

$$x=\frac{-25±95}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{-25±95}{10}$$

Now solve the equation $x=\frac{-25±95}{10}$ when $±$ is plus. Add $-25$ to $95$.

$$x=\frac{70}{10}$$

Divide $70$ by $10$.

$$x=7$$

Now solve the equation $x=\frac{-25±95}{10}$ when $±$ is minus. Subtract $95$ from $-25$.

$$x=-\frac{120}{10}$$

Divide $-120$ by $10$.

$$x=-12$$

The equation is now solved.

$$x=7$$ $$x=-12$$

Steps for Completing the Square

Variable $x$ cannot be equal to any of the values $-5,0$ since division by zero is not defined. Multiply both sides of the equation by $x\left(x+5\right)$.

$$84\left(x+5-x\right)=5x\left(x+5\right)$$

Combine $x$ and $-x$ to get $0$.

$$84\times 5=5x\left(x+5\right)$$

Multiply $84$ and $5$ to get $420$.

$$420=5x\left(x+5\right)$$

Use the distributive property to multiply $5x$ by $x+5$.

$$420=5x^{2}+25x$$

Swap sides so that all variable terms are on the left hand side.

$$5x^{2}+25x=420$$

Divide both sides by $5$.

$$\frac{5x^{2}+25x}{5}=\frac{420}{5}$$

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

$$x^{2}+\frac{25}{5}x=\frac{420}{5}$$

Divide $25$ by $5$.

$$x^{2}+5x=\frac{420}{5}$$

Divide $420$ by $5$.

$$x^{2}+5x=84$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=84+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=84+\frac{25}{4}$$

Add $84$ to $\frac{25}{4}$.

$$x^{2}+5x+\frac{25}{4}=\frac{361}{4}$$

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

$$\left(x+\frac{5}{2}\right)^{2}=\frac{361}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{361}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{19}{2}$$ $$x+\frac{5}{2}=-\frac{19}{2}$$

Subtract $\frac{5}{2}$ from both sides of the equation.

$$x=7$$ $$x=-12$$