Solve by Factoring square root of 10-x+x=4

$\sqrt{10 - x} + x = 4$

Step 1

Subtract $4$ from both sides of the equation.

$\sqrt{10 - x} + x - 4 = 0$

Step 2

Move all terms not containing $\sqrt{10 - x}$ to the right side of the equation.

Subtract $x$ from both sides of the equation.

$\sqrt{10 - x} - 4 = - x$

Add $4$ to both sides of the equation.

$\sqrt{10 - x} = - x + 4$

Step 3

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

${\sqrt{10 - x}}^{2} = {(- x + 4)}^{2}$

Step 4

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10 - x}$ as ${(10 - x)}^{\frac{1}{2}}$ .

${({(10 - x)}^{\frac{1}{2}})}^{2} = {(- x + 4)}^{2}$

Simplify the left side.

Simplify ${({(10 - x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(10 - x)}^{\frac{1}{2}})}^{2}$ .

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

${(10 - x)}^{\frac{1}{2} \cdot 2} = {(- x + 4)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(10 - x)}^{\frac{1}{2} \cdot 2} = {(- x + 4)}^{2}$

Rewrite the expression.

${(10 - x)}^{1} = {(- x + 4)}^{2}$

Simplify.

$10 - x = {(- x + 4)}^{2}$

Simplify the right side.

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

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

$10 - x = (- x + 4) (- x + 4)$

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

Apply the distributive property.

$10 - x = - x (- x + 4) + 4 (- x + 4)$

Apply the distributive property.

$10 - x = - x (- x) - x \cdot 4 + 4 (- x + 4)$

Apply the distributive property.

$10 - x = - x (- x) - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$10 - x = - 1 \cdot - 1 x \cdot x - x \cdot 4 + 4 (- x) + 4 \cdot 4$

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

Move $x$ .

$10 - x = - 1 \cdot - 1 (x \cdot x) - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $x$ by $x$ .

$10 - x = - 1 \cdot - 1 x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

Multiply $- 1$ by $- 1$ .

$10 - x = 1 x^{2} - x \cdot 4 + 4 (- x) + 4 \cdot 4$

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

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

Multiply $4$ by $- 1$ .

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

Multiply $- 1$ by $4$ .

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

Multiply $4$ by $4$ .

$10 - x = x^{2} - 4 x - 4 x + 16$

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

$10 - x = x^{2} - 8 x + 16$

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.

$x^{2} - 8 x + 16 = 10 - x$

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

Add $x$ to both sides of the equation.

$x^{2} - 8 x + 16 + x = 10$

Add $- 8 x$ and $x$ .

$x^{2} - 7 x + 16 = 10$

Subtract $10$ from both sides of the equation.

$x^{2} - 7 x + 16 - 10 = 0$

Subtract $10$ from $16$ .

$x^{2} - 7 x + 6 = 0$

Factor $x^{2} - 7 x + 6$ 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 $6$ and whose sum is $- 7$ .

$- 6, - 1$

Write the factored form using these integers.

$(x - 6) (x - 1) = 0$

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

$x - 6 = 0$

$x - 1 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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 $(x - 6) (x - 1) = 0$ true.

$x = 6, 1$

Step 6

Exclude the solutions that do not make $\sqrt{10 - x} + x - 4 = 0$ true.

$x = 1$

Solve by Factoring square root of 10-x+x=4

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