How to solve x-(5)/(x)=7

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-(5)/(x)=7 | Equatix

Question

$$x- \frac{ 5 }{ x } =7$$

Solve for x

$x = \frac{\sqrt{69} + 7}{2} \approx 7.653311931$
$x=\frac{7-\sqrt{69}}{2}\approx -0.653311931$

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$xx-5=7x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}-5=7x$$

Subtract $7x$ from both sides.

$$x^{2}-5-7x=0$$

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.

$$x^{2}-7x-5=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $-7$ for $b$, and $-5$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-5\right)}}{2}$$

Square $-7$.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-5\right)}}{2}$$

Multiply $-4$ times $-5$.

$$x=\frac{-\left(-7\right)±\sqrt{49+20}}{2}$$

Add $49$ to $20$.

$$x=\frac{-\left(-7\right)±\sqrt{69}}{2}$$

The opposite of $-7$ is $7$.

$$x=\frac{7±\sqrt{69}}{2}$$

Now solve the equation $x=\frac{7±\sqrt{69}}{2}$ when $±$ is plus. Add $7$ to $\sqrt{69}$.

$$x=\frac{\sqrt{69}+7}{2}$$

Now solve the equation $x=\frac{7±\sqrt{69}}{2}$ when $±$ is minus. Subtract $\sqrt{69}$ from $7$.

$$x=\frac{7-\sqrt{69}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{69}+7}{2}$$ $$x=\frac{7-\sqrt{69}}{2}$$

Steps for Completing the Square

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x$.

$$xx-5=7x$$

Multiply $x$ and $x$ to get $x^{2}$.

$$x^{2}-5=7x$$

Subtract $7x$ from both sides.

$$x^{2}-5-7x=0$$

Add $5$ to both sides. Anything plus zero gives itself.

$$x^{2}-7x=5$$

Divide $-7$, the coefficient of the $x$ term, by $2$ to get $-\frac{7}{2}$. Then add the square of $-\frac{7}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=5+\left(-\frac{7}{2}\right)^{2}$$

Square $-\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=5+\frac{49}{4}$$

Add $5$ to $\frac{49}{4}$.

$$x^{2}-7x+\frac{49}{4}=\frac{69}{4}$$

Factor $x^{2}-7x+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{69}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{69}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{\sqrt{69}}{2}$$ $$x-\frac{7}{2}=-\frac{\sqrt{69}}{2}$$

Add $\frac{7}{2}$ to both sides of the equation.

$$x=\frac{\sqrt{69}+7}{2}$$ $$x=\frac{7-\sqrt{69}}{2}$$

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Question

Solve for x

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

Steps for Completing the Square