Find the Inverse square root of 2x+5

$\sqrt{2 x} + 5$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Subtract $5$ from both sides of the equation.

$\sqrt{2 y} = x - 5$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

$2 y = (x - 5) (x - 5)$

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

Apply the distributive property.

$2 y = x (x - 5) - 5 (x - 5)$

Apply the distributive property.

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

Apply the distributive property.

$2 y = x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 5$ to the left of $x$ .

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

Multiply $- 5$ by $- 5$ .

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

Subtract $5 x$ from $- 5 x$ .

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

Divide each term in $2 y = x^{2} - 10 x + 25$ by $2$ and simplify.

Divide each term in $2 y = x^{2} - 10 x + 25$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Cancel the common factor of $- 10$ and $2$ .

Factor $2$ out of $- 10 x$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 5 x$ by $1$ .

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

Step 3

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

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

Step 4

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

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\sqrt{2 x} + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 5$ by $5$ .

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

To write $- 5 \sqrt{2 x}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Simplify terms.

Combine $- 5 \sqrt{2 x}$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Rewrite ${(\sqrt{2 x} + 5)}^{2}$ as $(\sqrt{2 x} + 5) (\sqrt{2 x} + 5)$ .

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

Expand $(\sqrt{2 x} + 5) (\sqrt{2 x} + 5)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt{2 x} \sqrt{2 x}$ .

Raise $\sqrt{2 x}$ to the power of $1$ .

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

Raise $\sqrt{2 x}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{2 x}}^{2}$ as $2 x$ .

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

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

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

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Move $5$ to the left of $\sqrt{2 x}$ .

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

Multiply $5$ by $5$ .

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

Add $5 \sqrt{2 x}$ and $5 \sqrt{2 x}$ .

$f^{- 1} (\sqrt{2 x} + 5) = \frac{2 x + 10 \sqrt{2 x} + 25 - 5 \sqrt{2 x} \cdot 2}{2} - 25 + \frac{25}{2}$

Multiply $2$ by $- 5$ .

$f^{- 1} (\sqrt{2 x} + 5) = \frac{2 x + 10 \sqrt{2 x} + 25 - 10 \sqrt{2 x}}{2} - 25 + \frac{25}{2}$

Subtract $10 \sqrt{2 x}$ from $10 \sqrt{2 x}$ .

$f^{- 1} (\sqrt{2 x} + 5) = \frac{2 x + 0 + 25}{2} - 25 + \frac{25}{2}$

Add $2 x$ and $0$ .

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

Simplify terms.

Combine the numerators over the common denominator.

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

Add $25$ and $25$ .

$f^{- 1} (\sqrt{2 x} + 5) = - 25 + \frac{2 x + 50}{2}$

Cancel the common factor of $2 x + 50$ and $2$ .

Factor $2$ out of $2 x$ .

$f^{- 1} (\sqrt{2 x} + 5) = - 25 + \frac{2 (x) + 50}{2}$

Factor $2$ out of $50$ .

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

Factor $2$ out of $2 (x) + 2 (25)$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $x + 25$ by $1$ .

$f^{- 1} (\sqrt{2 x} + 5) = - 25 + x + 25$

Combine the opposite terms in $- 25 + x + 25$ .

Add $- 25$ and $25$ .

$f^{- 1} (\sqrt{2 x} + 5) = x + 0$

Add $x$ and $0$ .

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{x^{2}}{2} - 5 x + \frac{25}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2}}{2} - 5 x + \frac{25}{2})} + 5$

Simplify each term.

To write $- 5 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2}}{2} - 5 x \cdot \frac{2}{2} + \frac{25}{2})} + 5$

Combine $- 5 x$ and $\frac{2}{2}$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2}}{2} + \frac{- 5 x \cdot 2}{2} + \frac{25}{2})} + 5$

Combine the numerators over the common denominator.

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2} - 5 x \cdot 2}{2} + \frac{25}{2})} + 5$

Combine the numerators over the common denominator.

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2} - 5 x \cdot 2 + 25}{2})} + 5$

Reorder factors in $x^{2} - 5 x \cdot 2 + 25$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2} - 5 \cdot (2 x) + 25}{2})} + 5$

Simplify the numerator.

Multiply $- 5$ by $2$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2} - 10 x + 25}{2})} + 5$

Factor using the perfect square rule.

Rewrite $25$ as $5^{2}$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2} - 10 x + 5^{2}}{2})} + 5$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 x = 2 \cdot x \cdot 5$

Rewrite the polynomial.

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{x^{2} - 2 \cdot x \cdot 5 + 5^{2}}{2})} + 5$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 5$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{2 (\frac{{(x - 5)}^{2}}{2})} + 5$

Combine $2$ and $\frac{{(x - 5)}^{2}}{2}$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{\frac{2 {(x - 5)}^{2}}{2}} + 5$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{2 {(x - 5)}^{2}}{2}$ by cancelling the common factors.

Cancel the common factor.

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{\frac{2 {(x - 5)}^{2}}{2}} + 5$

Rewrite the expression.

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

Divide ${(x - 5)}^{2}$ by $1$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = \sqrt{{(x - 5)}^{2}} + 5$

Pull terms out from under the radical, assuming positive real numbers.

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = x - 5 + 5$

Combine the opposite terms in $x - 5 + 5$ .

Add $- 5$ and $5$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = x + 0$

Add $x$ and $0$ .

$f (\frac{x^{2}}{2} - 5 x + \frac{25}{2}) = x$

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

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

Find the Inverse square root of 2x+5

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