Solve the Rational Equation for x square root of 2x+5-1=x

$\sqrt{2 x + 5} - 1 = x$

Add $1$ to both sides of the equation.

$\sqrt{2 x + 5} = x + 1$

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

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x + 5}$ as ${(2 x + 5)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${({(2 x + 5)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 x + 5)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

${(2 x + 5)}^{1} = {(x + 1)}^{2}$

Simplify.

$2 x + 5 = {(x + 1)}^{2}$

Simplify the right side.

Simplify ${(x + 1)}^{2}$ .

Rewrite ${(x + 1)}^{2}$ as $(x + 1) (x + 1)$ .

$2 x + 5 = (x + 1) (x + 1)$

Expand $(x + 1) (x + 1)$ using the FOIL Method.

Apply the distributive property.

$2 x + 5 = x (x + 1) + 1 (x + 1)$

Apply the distributive property.

$2 x + 5 = x \cdot x + x \cdot 1 + 1 (x + 1)$

Apply the distributive property.

$2 x + 5 = x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$2 x + 5 = x^{2} + x \cdot 1 + 1 x + 1 \cdot 1$

Multiply $x$ by $1$ .

$2 x + 5 = x^{2} + x + 1 x + 1 \cdot 1$

Multiply $x$ by $1$ .

$2 x + 5 = x^{2} + x + x + 1 \cdot 1$

Multiply $1$ by $1$ .

$2 x + 5 = x^{2} + x + x + 1$

Add $x$ and $x$ .

$2 x + 5 = x^{2} + 2 x + 1$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 2 x + 1 = 2 x + 5$

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

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

Combine the opposite terms in $x^{2} + 2 x + 1 - 2 x$ .

Subtract $2 x$ from $2 x$ .

$x^{2} + 0 + 1 = 5$

Add $x^{2}$ and $0$ .

$x^{2} + 1 = 5$

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

Subtract $1$ from both sides of the equation.

$x^{2} = 5 - 1$

Subtract $1$ from $5$ .

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

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

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Exclude the solutions that do not make $\sqrt{2 x + 5} - 1 = x$ true.

$x = 2$

Solve the Rational Equation for x square root of 2x+5-1=x

Related Questions

Support LCMGCF.com by disabling your adblocker.