Solve 15/x - (11x + 5)/(x ^ 2) = - 1 | 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

$$\frac{15}{x}-\frac{11x+5}{x^{2}}=-1$$

Solve for x

$x=-5$
$x=1$

Steps Using Factoring

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x^{2}$, the least common multiple of $x,x^{2}$.

$$x\times 15-\left(11x+5\right)=-x^{2}$$

To find the opposite of $11x+5$, find the opposite of each term.

$$x\times 15-11x-5=-x^{2}$$

Combine $x\times 15$ and $-11x$ to get $4x$.

$$4x-5=-x^{2}$$

Add $x^{2}$ to both sides.

$$4x-5+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+4x-5=0$$

To solve the equation, factor $x^{2}+4x-5$ using formula $x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right)$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=-5$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1$$ $$b=5$$

Rewrite factored expression $\left(x+a\right)\left(x+b\right)$ using the obtained values.

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve $x-1=0$ and $x+5=0$.

$$x=1$$ $$x=-5$$

Steps Using Factoring By Grouping

Variable $x$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $x^{2}$, the least common multiple of $x,x^{2}$.

$$x\times 15-\left(11x+5\right)=-x^{2}$$

To find the opposite of $11x+5$, find the opposite of each term.

$$x\times 15-11x-5=-x^{2}$$

Combine $x\times 15$ and $-11x$ to get $4x$.

$$4x-5=-x^{2}$$

Add $x^{2}$ to both sides.

$$4x-5+x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}+4x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $x^{2}+ax+bx-5$. To find $a$ and $b$, set up a system to be solved.

$$a+b=4$$ $$ab=1\left(-5\right)=-5$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1$$ $$b=5$$

Rewrite $x^{2}+4x-5$ as $\left(x^{2}-x\right)+\left(5x-5\right)$.

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

Factor out $x$ in the first and $5$ in the second group.

$$x\left(x-1\right)+5\left(x-1\right)$$

Factor out common term $x-1$ by using distributive property.

$$\left(x-1\right)\left(x+5\right)$$

To find equation solutions, solve $x-1=0$ and $x+5=0$.

$$x=1$$ $$x=-5$$

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 $x^{2}$, the least common multiple of $x,x^{2}$.

$$x\times 15-\left(11x+5\right)=-x^{2}$$

To find the opposite of $11x+5$, find the opposite of each term.

$$x\times 15-11x-5=-x^{2}$$

Combine $x\times 15$ and $-11x$ to get $4x$.

$$4x-5=-x^{2}$$

Add $x^{2}$ to both sides.

$$4x-5+x^{2}=0$$

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.

$$x^{2}+4x-5=0$$

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

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

Square $4$.

$$x=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply $-4$ times $-5$.

$$x=\frac{-4±\sqrt{16+20}}{2}$$

Add $16$ to $20$.

$$x=\frac{-4±\sqrt{36}}{2}$$

Take the square root of $36$.

$$x=\frac{-4±6}{2}$$

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

$$x=\frac{2}{2}$$

Divide $2$ by $2$.

$$x=1$$

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

$$x=-\frac{10}{2}$$

Divide $-10$ by $2$.

$$x=-5$$

The equation is now solved.

$$x=1$$ $$x=-5$$

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 $x^{2}$, the least common multiple of $x,x^{2}$.

$$x\times 15-\left(11x+5\right)=-x^{2}$$

To find the opposite of $11x+5$, find the opposite of each term.

$$x\times 15-11x-5=-x^{2}$$

Combine $x\times 15$ and $-11x$ to get $4x$.

$$4x-5=-x^{2}$$

Add $x^{2}$ to both sides.

$$4x-5+x^{2}=0$$

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

$$4x+x^{2}=5$$

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

$$x^{2}+4x=5$$

Divide $4$, the coefficient of the $x$ term, by $2$ to get $2$. Then add the square of $2$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=5+2^{2}$$

Square $2$.

$$x^{2}+4x+4=5+4$$

Add $5$ to $4$.

$$x^{2}+4x+4=9$$

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

$$\left(x+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$x+2=3$$ $$x+2=-3$$

Subtract $2$ from both sides of the equation.

$$x=1$$ $$x=-5$$