Algebra Examples

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

Step 1

Rewrite the equation as $y^{2} - 5 y = x$ .

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

Step 2

Subtract $x$ from both sides of the equation.

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

Step 3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4

Substitute the values $a = 1$ , $b = - 5$ , and $c = - x$ into the quadratic formula and solve for $y$ .

$\frac{5 \pm \sqrt{{(- 5)}^{2} - 4 \cdot (1 \cdot (- x))}}{2 \cdot 1}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot (- 1 x)}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 5.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot (- 1 x)}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $- 1$ .

$y = \frac{5 \pm \sqrt{25 + 4 x}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$y = \frac{5 \pm \sqrt{25 + 4 x}}{2}$

Step 6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot (- 1 x)}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 6.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot (- 1 x)}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $- 1$ .

$y = \frac{5 \pm \sqrt{25 + 4 x}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$y = \frac{5 \pm \sqrt{25 + 4 x}}{2}$

Step 6.3

Change the $\pm$ to $+$ .

$y = \frac{5 + \sqrt{25 + 4 x}}{2}$

Step 7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Raise $- 5$ to the power of $2$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 1 \cdot (- 1 x)}}{2 \cdot 1}$

Step 7.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

$y = \frac{5 \pm \sqrt{25 - 4 \cdot (- 1 x)}}{2 \cdot 1}$

Step 7.1.2.2

Multiply $- 4$ by $- 1$ .

$y = \frac{5 \pm \sqrt{25 + 4 x}}{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$y = \frac{5 \pm \sqrt{25 + 4 x}}{2}$

Step 7.3

Change the $\pm$ to $-$ .

$y = \frac{5 - \sqrt{25 + 4 x}}{2}$

Step 8

The final answer is the combination of both solutions.

$y = \frac{5 + \sqrt{25 + 4 x}}{2}$

$y = \frac{5 - \sqrt{25 + 4 x}}{2}$

Step 9

Interchange the variables. Create an equation for each expression.

$x = \frac{5 + \sqrt{25 + 4 y}}{2}, x = \frac{5 - \sqrt{25 + 4 y}}{2}$

Step 10

Solve for $y$ .

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Step 10.1

Rewrite the equation as $\frac{5 + \sqrt{25 + 4 y}}{2} = x$ .

$\frac{5 + \sqrt{25 + 4 y}}{2} = x$

Step 10.2

Multiply both sides by $2$ .

$\frac{5 + \sqrt{25 + 4 y}}{2} \cdot 2 = x \cdot 2$

Step 10.3

Simplify.

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Step 10.3.1

Simplify the left side.

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Step 10.3.1.1

Simplify $\frac{5 + \sqrt{25 + 4 y}}{2} \cdot 2$ .

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Step 10.3.1.1.1

Cancel the common factor of $2$ .

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Step 10.3.1.1.1.1

Cancel the common factor.

$\frac{5 + \sqrt{25 + 4 y}}{2} \cdot 2 = x \cdot 2$

Step 10.3.1.1.1.2

Rewrite the expression.

$5 + \sqrt{25 + 4 y} = x \cdot 2$

Step 10.3.1.1.2

Reorder $25$ and $4 y$ .

$5 + \sqrt{4 y + 25} = x \cdot 2$

Step 10.3.2

Simplify the right side.

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Step 10.3.2.1

Move $2$ to the left of $x$ .

$5 + \sqrt{4 y + 25} = 2 x$

Step 10.4

Solve for $y$ .

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Step 10.4.1

Subtract $5$ from both sides of the equation.

$\sqrt{4 y + 25} = 2 x - 5$

Step 10.4.2

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{4 y + 25}}^{2} = {(2 x - 5)}^{2}$

Step 10.4.3

Simplify each side of the equation.

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Step 10.4.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{4 y + 25}$ as ${(4 y + 25)}^{\frac{1}{2}}$ .

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

Step 10.4.3.2

Simplify the left side.

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Step 10.4.3.2.1

Simplify ${({(4 y + 25)}^{\frac{1}{2}})}^{2}$ .

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Step 10.4.3.2.1.1

Multiply the exponents in ${({(4 y + 25)}^{\frac{1}{2}})}^{2}$ .

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Step 10.4.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(4 y + 25)}^{\frac{1}{2} \cdot 2} = {(2 x - 5)}^{2}$

Step 10.4.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 10.4.3.2.1.1.2.1

Cancel the common factor.

${(4 y + 25)}^{\frac{1}{2} \cdot 2} = {(2 x - 5)}^{2}$

Step 10.4.3.2.1.1.2.2

Rewrite the expression.

${(4 y + 25)}^{1} = {(2 x - 5)}^{2}$

Step 10.4.3.2.1.2

Simplify.

$4 y + 25 = {(2 x - 5)}^{2}$

Step 10.4.3.3

Simplify the right side.

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Step 10.4.3.3.1

Simplify ${(2 x - 5)}^{2}$ .

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Step 10.4.3.3.1.1

Rewrite ${(2 x - 5)}^{2}$ as $(2 x - 5) (2 x - 5)$ .

$4 y + 25 = (2 x - 5) (2 x - 5)$

Step 10.4.3.3.1.2

Expand $(2 x - 5) (2 x - 5)$ using the FOIL Method.

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Step 10.4.3.3.1.2.1

Apply the distributive property.

$4 y + 25 = 2 x (2 x - 5) - 5 (2 x - 5)$

Step 10.4.3.3.1.2.2

Apply the distributive property.

$4 y + 25 = 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x - 5)$

Step 10.4.3.3.1.2.3

Apply the distributive property.

$4 y + 25 = 2 x (2 x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Step 10.4.3.3.1.3

Simplify and combine like terms.

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Step 10.4.3.3.1.3.1

Simplify each term.

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Step 10.4.3.3.1.3.1.1

Rewrite using the commutative property of multiplication.

$4 y + 25 = 2 \cdot 2 x \cdot x + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Step 10.4.3.3.1.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 10.4.3.3.1.3.1.2.1

Move $x$ .

$4 y + 25 = 2 \cdot 2 (x \cdot x) + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Step 10.4.3.3.1.3.1.2.2

Multiply $x$ by $x$ .

$4 y + 25 = 2 \cdot 2 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Step 10.4.3.3.1.3.1.3

Multiply $2$ by $2$ .

$4 y + 25 = 4 x^{2} + 2 x \cdot - 5 - 5 (2 x) - 5 \cdot - 5$

Step 10.4.3.3.1.3.1.4

Multiply $- 5$ by $2$ .

$4 y + 25 = 4 x^{2} - 10 x - 5 (2 x) - 5 \cdot - 5$

Step 10.4.3.3.1.3.1.5

Multiply $2$ by $- 5$ .

$4 y + 25 = 4 x^{2} - 10 x - 10 x - 5 \cdot - 5$

Step 10.4.3.3.1.3.1.6

Multiply $- 5$ by $- 5$ .

$4 y + 25 = 4 x^{2} - 10 x - 10 x + 25$

Step 10.4.3.3.1.3.2

Subtract $10 x$ from $- 10 x$ .

$4 y + 25 = 4 x^{2} - 20 x + 25$

Step 10.4.4

Solve for $y$ .

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Step 10.4.4.1

Move all terms not containing $y$ to the right side of the equation.

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Step 10.4.4.1.1

Subtract $25$ from both sides of the equation.

$4 y = 4 x^{2} - 20 x + 25 - 25$

Step 10.4.4.1.2

Combine the opposite terms in $4 x^{2} - 20 x + 25 - 25$ .

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Step 10.4.4.1.2.1

Subtract $25$ from $25$ .

$4 y = 4 x^{2} - 20 x + 0$

Step 10.4.4.1.2.2

Add $4 x^{2} - 20 x$ and $0$ .

$4 y = 4 x^{2} - 20 x$

Step 10.4.4.2

Divide each term in $4 y = 4 x^{2} - 20 x$ by $4$ and simplify.

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Step 10.4.4.2.1

Divide each term in $4 y = 4 x^{2} - 20 x$ by $4$ .

$\frac{4 y}{4} = \frac{4 x^{2}}{4} + \frac{- 20 x}{4}$

Step 10.4.4.2.2

Simplify the left side.

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Step 10.4.4.2.2.1

Cancel the common factor of $4$ .

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Step 10.4.4.2.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{4 x^{2}}{4} + \frac{- 20 x}{4}$

Step 10.4.4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{4 x^{2}}{4} + \frac{- 20 x}{4}$

Step 10.4.4.2.3

Simplify the right side.

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Step 10.4.4.2.3.1

Simplify each term.

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Step 10.4.4.2.3.1.1

Cancel the common factor of $4$ .

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Step 10.4.4.2.3.1.1.1

Cancel the common factor.

$y = \frac{4 x^{2}}{4} + \frac{- 20 x}{4}$

Step 10.4.4.2.3.1.1.2

Divide $x^{2}$ by $1$ .

$y = x^{2} + \frac{- 20 x}{4}$

Step 10.4.4.2.3.1.2

Cancel the common factor of $- 20$ and $4$ .

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Step 10.4.4.2.3.1.2.1

Factor $4$ out of $- 20 x$ .

$y = x^{2} + \frac{4 (- 5 x)}{4}$

Step 10.4.4.2.3.1.2.2

Cancel the common factors.

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Step 10.4.4.2.3.1.2.2.1

Factor $4$ out of $4$ .

$y = x^{2} + \frac{4 (- 5 x)}{4 (1)}$

Step 10.4.4.2.3.1.2.2.2

Cancel the common factor.

$y = x^{2} + \frac{4 (- 5 x)}{4 \cdot 1}$

Step 10.4.4.2.3.1.2.2.3

Rewrite the expression.

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

Step 10.4.4.2.3.1.2.2.4

Divide $- 5 x$ by $1$ .

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

Step 11

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = x^{2} - 5 x$

Step 12

Verify if $f^{- 1} (x) = x^{2} - 5 x$ is the inverse of $f (x) = \frac{5 + \sqrt{25 + 4 x}}{2}, \frac{5 - \sqrt{25 + 4 x}}{2}$ .

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Step 12.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \frac{5 + \sqrt{25 + 4 x}}{2}, \frac{5 - \sqrt{25 + 4 x}}{2}$ and $f^{- 1} (x) = x^{2} - 5 x$ and compare them.

Step 12.2

Find the range of $f (x) = \frac{5 + \sqrt{25 + 4 x}}{2}, \frac{5 - \sqrt{25 + 4 x}}{2}$ .

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Step 12.2.1

Find the range of $\frac{5 + \sqrt{25 + 4 x}}{2}$ .

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Step 12.2.1.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[\frac{5}{2}, \infty)$

Step 12.2.2

Find the range of $\frac{5 - \sqrt{25 + 4 x}}{2}$ .

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Step 12.2.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \frac{5}{2}]$

Step 12.2.3

Find the union of $[\frac{5}{2}, \infty), (- \infty, \frac{5}{2}]$ .

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Step 12.2.3.1

The union consists of all of the elements that are contained in each interval.

$(- \infty, \infty)$

Step 12.3

Find the domain of $\frac{5 + \sqrt{25 + 4 x}}{2}$ .

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Step 12.3.1

Set the radicand in $\sqrt{25 + 4 x}$ greater than or equal to $0$ to find where the expression is defined.

$25 + 4 x \geq 0$

Step 12.3.2

Solve for $x$ .

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Step 12.3.2.1

Subtract $25$ from both sides of the inequality.

$4 x \geq - 25$

Step 12.3.2.2

Divide each term in $4 x \geq - 25$ by $4$ and simplify.

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Step 12.3.2.2.1

Divide each term in $4 x \geq - 25$ by $4$ .

$\frac{4 x}{4} \geq \frac{- 25}{4}$

Step 12.3.2.2.2

Simplify the left side.

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Step 12.3.2.2.2.1

Cancel the common factor of $4$ .

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Step 12.3.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} \geq \frac{- 25}{4}$

Step 12.3.2.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{- 25}{4}$

Step 12.3.2.2.3

Simplify the right side.

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Step 12.3.2.2.3.1

Move the negative in front of the fraction.

$x \geq - \frac{25}{4}$

Step 12.3.3

The domain is all values of $x$ that make the expression defined.

$[- \frac{25}{4}, \infty)$

Step 12.4

Since the domain of $f^{- 1} (x) = x^{2} - 5 x$ is the range of $f (x) = \frac{5 + \sqrt{25 + 4 x}}{2}, \frac{5 - \sqrt{25 + 4 x}}{2}$ and the range of $f^{- 1} (x) = x^{2} - 5 x$ is the domain of $f (x) = \frac{5 + \sqrt{25 + 4 x}}{2}, \frac{5 - \sqrt{25 + 4 x}}{2}$ , then $f^{- 1} (x) = x^{2} - 5 x$ is the inverse of $f (x) = \frac{5 + \sqrt{25 + 4 x}}{2}, \frac{5 - \sqrt{25 + 4 x}}{2}$ .

$f^{- 1} (x) = x^{2} - 5 x$

Step 13