Solve 3x ^ - 2 + 5 = 8x ^ - 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

$$3x^{-2}+5=8x^{-1}$$

Solve for x

$x=\frac{3}{5}=0.6$
$x=1$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Subtract $8x^{-1}$ from both sides.

$$3x^{-2}+5-8x^{-1}=0$$

Reorder the terms.

$$5-8\times \frac{1}{x}+3x^{-2}=0$$

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

$$x\times 5-8+3x^{-2}x=0$$

To multiply powers of the same base, add their exponents. Add $-2$ and $1$ to get $-1$.

$$x\times 5-8+3x^{-1}=0$$

Reorder the terms.

$$5x-8+3\times \frac{1}{x}=0$$

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

$$5xx+x\left(-8\right)+3\times 1=0$$

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

$$5x^{2}+x\left(-8\right)+3\times 1=0$$

Multiply $3$ and $1$ to get $3$.

$$5x^{2}+x\left(-8\right)+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $5x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-8$$ $$ab=5\times 3=15$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $15$.

$$-1,-15$$ $$-3,-5$$

Calculate the sum for each pair.

$$-1-15=-16$$ $$-3-5=-8$$

The solution is the pair that gives sum $-8$.

$$a=-5$$ $$b=-3$$

Rewrite $5x^{2}-8x+3$ as $\left(5x^{2}-5x\right)+\left(-3x+3\right)$.

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

Factor out $5x$ in the first and $-3$ in the second group.

$$5x\left(x-1\right)-3\left(x-1\right)$$

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

$$\left(x-1\right)\left(5x-3\right)$$

To find equation solutions, solve $x-1=0$ and $5x-3=0$.

$$x=1$$ $$x=\frac{3}{5}$$

Steps Using the Quadratic Formula

Subtract $8x^{-1}$ from both sides.

$$3x^{-2}+5-8x^{-1}=0$$

Reorder the terms.

$$5-8\times \frac{1}{x}+3x^{-2}=0$$

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

$$x\times 5-8+3x^{-2}x=0$$

To multiply powers of the same base, add their exponents. Add $-2$ and $1$ to get $-1$.

$$x\times 5-8+3x^{-1}=0$$

Reorder the terms.

$$5x-8+3\times \frac{1}{x}=0$$

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

$$5xx+x\left(-8\right)+3\times 1=0$$

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

$$5x^{2}+x\left(-8\right)+3\times 1=0$$

Multiply $3$ and $1$ to get $3$.

$$5x^{2}+x\left(-8\right)+3=0$$

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.

$$5x^{2}-8x+3=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $5$ for $a$, $-8$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\times 3}}{2\times 5}$$

Square $-8$.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\times 3}}{2\times 5}$$

Multiply $-4$ times $5$.

$$x=\frac{-\left(-8\right)±\sqrt{64-20\times 3}}{2\times 5}$$

Multiply $-20$ times $3$.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2\times 5}$$

Add $64$ to $-60$.

$$x=\frac{-\left(-8\right)±\sqrt{4}}{2\times 5}$$

Take the square root of $4$.

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

The opposite of $-8$ is $8$.

$$x=\frac{8±2}{2\times 5}$$

Multiply $2$ times $5$.

$$x=\frac{8±2}{10}$$

Now solve the equation $x=\frac{8±2}{10}$ when $±$ is plus. Add $8$ to $2$.

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

Divide $10$ by $10$.

$$x=1$$

Now solve the equation $x=\frac{8±2}{10}$ when $±$ is minus. Subtract $2$ from $8$.

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

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

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

The equation is now solved.

$$x=1$$ $$x=\frac{3}{5}$$

Steps for Completing the Square

Subtract $8x^{-1}$ from both sides.

$$3x^{-2}+5-8x^{-1}=0$$

Subtract $5$ from both sides. Anything subtracted from zero gives its negation.

$$3x^{-2}-8x^{-1}=-5$$

Reorder the terms.

$$-8\times \frac{1}{x}+3x^{-2}=-5$$

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

$$-8+3x^{-2}x=-5x$$

To multiply powers of the same base, add their exponents. Add $-2$ and $1$ to get $-1$.

$$-8+3x^{-1}=-5x$$

Add $5x$ to both sides.

$$-8+3x^{-1}+5x=0$$

Add $8$ to both sides. Anything plus zero gives itself.

$$3x^{-1}+5x=8$$

Reorder the terms.

$$5x+3\times \frac{1}{x}=8$$

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

$$5xx+3\times 1=8x$$

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

$$5x^{2}+3\times 1=8x$$

Multiply $3$ and $1$ to get $3$.

$$5x^{2}+3=8x$$

Subtract $8x$ from both sides.

$$5x^{2}+3-8x=0$$

Subtract $3$ from both sides. Anything subtracted from zero gives its negation.

$$5x^{2}-8x=-3$$

Divide both sides by $5$.

$$\frac{5x^{2}-8x}{5}=-\frac{3}{5}$$

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

$$x^{2}-\frac{8}{5}x=-\frac{3}{5}$$

Divide $-\frac{8}{5}$, the coefficient of the $x$ term, by $2$ to get $-\frac{4}{5}$. Then add the square of $-\frac{4}{5}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{3}{5}+\left(-\frac{4}{5}\right)^{2}$$

Square $-\frac{4}{5}$ by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=-\frac{3}{5}+\frac{16}{25}$$

Add $-\frac{3}{5}$ to $\frac{16}{25}$ by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{1}{25}$$

Factor $x^{2}-\frac{8}{5}x+\frac{16}{25}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x-\frac{4}{5}\right)^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Simplify.

$$x-\frac{4}{5}=\frac{1}{5}$$ $$x-\frac{4}{5}=-\frac{1}{5}$$

Add $\frac{4}{5}$ to both sides of the equation.

$$x=1$$ $$x=\frac{3}{5}$$