Solve by Factoring square root of 41-x=x+1

$\sqrt{41 - x} = x + 1$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x$ from both sides of the equation.

$\sqrt{41 - x} - x = 1$

Subtract $1$ from both sides of the equation.

$\sqrt{41 - x} - x - 1 = 0$

Step 2

Move all terms not containing $\sqrt{41 - x}$ to the right side of the equation.

Add $x$ to both sides of the equation.

$\sqrt{41 - x} - 1 = x$

Add $1$ to both sides of the equation.

$\sqrt{41 - x} = x + 1$

Step 3

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

${\sqrt{41 - x}}^{2} = {(x + 1)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41 - x}$ as ${(41 - x)}^{\frac{1}{2}}$ .

${({(41 - x)}^{\frac{1}{2}})}^{2} = {(x + 1)}^{2}$

Simplify the left side.

Simplify ${({(41 - x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(41 - x)}^{\frac{1}{2}})}^{2}$ .

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

${(41 - x)}^{\frac{1}{2} \cdot 2} = {(x + 1)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(41 - x)}^{\frac{1}{2} \cdot 2} = {(x + 1)}^{2}$

Rewrite the expression.

${(41 - x)}^{1} = {(x + 1)}^{2}$

Simplify.

$41 - x = {(x + 1)}^{2}$

Simplify the right side.

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

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

$41 - x = (x + 1) (x + 1)$

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

Apply the distributive property.

$41 - x = x (x + 1) + 1 (x + 1)$

Apply the distributive property.

$41 - x = x \cdot x + x \cdot 1 + 1 (x + 1)$

Apply the distributive property.

$41 - x = x \cdot x + x \cdot 1 + 1 x + 1 \cdot 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$41 - x = x^{2} + x \cdot 1 + 1 x + 1 \cdot 1$

Multiply $x$ by $1$ .

$41 - x = x^{2} + x + 1 x + 1 \cdot 1$

Multiply $x$ by $1$ .

$41 - x = x^{2} + x + x + 1 \cdot 1$

Multiply $1$ by $1$ .

$41 - x = x^{2} + x + x + 1$

Add $x$ and $x$ .

$41 - x = x^{2} + 2 x + 1$

Step 5

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.

$x^{2} + 2 x + 1 = 41 - x$

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

Add $x$ to both sides of the equation.

$x^{2} + 2 x + 1 + x = 41$

Add $2 x$ and $x$ .

$x^{2} + 3 x + 1 = 41$

Subtract $41$ from both sides of the equation.

$x^{2} + 3 x + 1 - 41 = 0$

Subtract $41$ from $1$ .

$x^{2} + 3 x - 40 = 0$

Factor $x^{2} + 3 x - 40$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 40$ and whose sum is $3$ .

$- 5, 8$

Write the factored form using these integers.

$(x - 5) (x + 8) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x + 8 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $(x - 5) (x + 8) = 0$ true.

$x = 5, - 8$

Step 6

Exclude the solutions that do not make $\sqrt{41 - x} - x - 1 = 0$ true.

$x = 5$

Solve by Factoring square root of 41-x=x+1

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