Solve by Factoring square root of 5-x- square root of x=1

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

Step 1

Subtract $1$ from both sides of the equation.

$\sqrt{5 - x} - \sqrt{x} - 1 = 0$

Step 2

Move all terms not containing $\sqrt{5 - x}$ to the right side of the equation.

Add $\sqrt{x}$ to both sides of the equation.

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

Add $1$ to both sides of the equation.

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

Step 3

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

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

Step 4

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

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

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

Expand $(\sqrt{x} + 1) (\sqrt{x} + 1)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$5 - x = \sqrt{x} \sqrt{x} + \sqrt{x} \cdot 1 + 1 (\sqrt{x} + 1)$

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

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

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

$5 - x = {\sqrt{x}}^{1} \sqrt{x} + \sqrt{x} \cdot 1 + 1 \sqrt{x} + 1 \cdot 1$

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

$5 - x = {\sqrt{x}}^{1} {\sqrt{x}}^{1} + \sqrt{x} \cdot 1 + 1 \sqrt{x} + 1 \cdot 1$

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

$5 - x = {\sqrt{x}}^{1 + 1} + \sqrt{x} \cdot 1 + 1 \sqrt{x} + 1 \cdot 1$

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$5 - x = x^{1} + \sqrt{x} \cdot 1 + 1 \sqrt{x} + 1 \cdot 1$

Simplify.

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

Multiply $\sqrt{x}$ by $1$ .

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

Multiply $\sqrt{x}$ by $1$ .

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

Multiply $1$ by $1$ .

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

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

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

Step 5

Solve for $2 \sqrt{x}$ .

Rewrite the equation as $x + 2 \sqrt{x} + 1 = 5 - x$ .

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

Move all terms not containing $2 \sqrt{x}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

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

Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $5$ .

$2 \sqrt{x} = - x - x + 4$

Subtract $x$ from $- x$ .

$2 \sqrt{x} = - 2 x + 4$

Step 6

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

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

Step 7

Simplify each side of the equation.

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

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

Simplify the left side.

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

Apply the product rule to $2 x^{\frac{1}{2}}$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(- 2 x + 4)}^{2}$ .

Rewrite ${(- 2 x + 4)}^{2}$ as $(- 2 x + 4) (- 2 x + 4)$ .

$4 x = (- 2 x + 4) (- 2 x + 4)$

Expand $(- 2 x + 4) (- 2 x + 4)$ using the FOIL Method.

Apply the distributive property.

$4 x = - 2 x (- 2 x + 4) + 4 (- 2 x + 4)$

Apply the distributive property.

$4 x = - 2 x (- 2 x) - 2 x \cdot 4 + 4 (- 2 x + 4)$

Apply the distributive property.

$4 x = - 2 x (- 2 x) - 2 x \cdot 4 + 4 (- 2 x) + 4 \cdot 4$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$4 x = - 2 \cdot - 2 x \cdot x - 2 x \cdot 4 + 4 (- 2 x) + 4 \cdot 4$

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

Move $x$ .

$4 x = - 2 \cdot - 2 (x \cdot x) - 2 x \cdot 4 + 4 (- 2 x) + 4 \cdot 4$

Multiply $x$ by $x$ .

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

Multiply $- 2$ by $- 2$ .

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

Multiply $4$ by $- 2$ .

$4 x = 4 x^{2} - 8 x + 4 (- 2 x) + 4 \cdot 4$

Multiply $- 2$ by $4$ .

$4 x = 4 x^{2} - 8 x - 8 x + 4 \cdot 4$

Multiply $4$ by $4$ .

$4 x = 4 x^{2} - 8 x - 8 x + 16$

Subtract $8 x$ from $- 8 x$ .

$4 x = 4 x^{2} - 16 x + 16$

Step 8

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.

$4 x^{2} - 16 x + 16 = 4 x$

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

Subtract $4 x$ from both sides of the equation.

$4 x^{2} - 16 x + 16 - 4 x = 0$

Subtract $4 x$ from $- 16 x$ .

$4 x^{2} - 20 x + 16 = 0$

Factor the left side of the equation.

Factor $4$ out of $4 x^{2} - 20 x + 16$ .

Factor $4$ out of $4 x^{2}$ .

$4 (x^{2}) - 20 x + 16 = 0$

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

$4 (x^{2}) + 4 (- 5 x) + 16 = 0$

Factor $4$ out of $16$ .

$4 x^{2} + 4 (- 5 x) + 4 \cdot 4 = 0$

Factor $4$ out of $4 x^{2} + 4 (- 5 x)$ .

$4 (x^{2} - 5 x) + 4 \cdot 4 = 0$

Factor $4$ out of $4 (x^{2} - 5 x) + 4 \cdot 4$ .

$4 (x^{2} - 5 x + 4) = 0$

Factor.

Factor $x^{2} - 5 x + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$4 ((x - 4) (x - 1)) = 0$

Remove unnecessary parentheses.

$4 (x - 4) (x - 1) = 0$

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

$x - 4 = 0$

$x - 1 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $4 (x - 4) (x - 1) = 0$ true.

$x = 4, 1$

Step 9

Exclude the solutions that do not make $\sqrt{5 - x} - \sqrt{x} - 1 = 0$ true.

$x = 1$

Solve by Factoring square root of 5-x- square root of x=1

Related Questions

Support LCMGCF.com by disabling your adblocker.