Solve for x square root of 11-2x = square root of x^2+4x+4

$\sqrt{11 - 2 x} = \sqrt{x^{2} + 4 x + 4}$

Step 1

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

${\sqrt{11 - 2 x}}^{2} = {\sqrt{x^{2} + 4 x + 4}}^{2}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{11 - 2 x}$ as ${(11 - 2 x)}^{\frac{1}{2}}$ .

${({(11 - 2 x)}^{\frac{1}{2}})}^{2} = {\sqrt{x^{2} + 4 x + 4}}^{2}$

Simplify the left side.

Simplify ${({(11 - 2 x)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(11 - 2 x)}^{\frac{1}{2}})}^{2}$ .

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

${(11 - 2 x)}^{\frac{1}{2} \cdot 2} = {\sqrt{x^{2} + 4 x + 4}}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(11 - 2 x)}^{\frac{1}{2} \cdot 2} = {\sqrt{x^{2} + 4 x + 4}}^{2}$

Rewrite the expression.

${(11 - 2 x)}^{1} = {\sqrt{x^{2} + 4 x + 4}}^{2}$

Simplify.

$11 - 2 x = {\sqrt{x^{2} + 4 x + 4}}^{2}$

Simplify the right side.

Simplify ${\sqrt{x^{2} + 4 x + 4}}^{2}$ .

Factor using the perfect square rule.

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

$11 - 2 x = {\sqrt{x^{2} + 4 x + 2^{2}}}^{2}$

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

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$11 - 2 x = {\sqrt{x^{2} + 2 \cdot x \cdot 2 + 2^{2}}}^{2}$

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

$11 - 2 x = {\sqrt{{(x + 2)}^{2}}}^{2}$

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

$11 - 2 x = {(x + 2)}^{2}$

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

$11 - 2 x = (x + 2) (x + 2)$

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

Apply the distributive property.

$11 - 2 x = x (x + 2) + 2 (x + 2)$

Apply the distributive property.

$11 - 2 x = x \cdot x + x \cdot 2 + 2 (x + 2)$

Apply the distributive property.

$11 - 2 x = x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$11 - 2 x = x^{2} + x \cdot 2 + 2 x + 2 \cdot 2$

Move $2$ to the left of $x$ .

$11 - 2 x = x^{2} + 2 \cdot x + 2 x + 2 \cdot 2$

Multiply $2$ by $2$ .

$11 - 2 x = x^{2} + 2 x + 2 x + 4$

Add $2 x$ and $2 x$ .

$11 - 2 x = x^{2} + 4 x + 4$

Step 3

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.

$x^{2} + 4 x + 4 = 11 - 2 x$

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

Add $2 x$ to both sides of the equation.

$x^{2} + 4 x + 4 + 2 x = 11$

Add $4 x$ and $2 x$ .

$x^{2} + 6 x + 4 = 11$

Subtract $11$ from both sides of the equation.

$x^{2} + 6 x + 4 - 11 = 0$

Subtract $11$ from $4$ .

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

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

$- 1, 7$

Write the factored form using these integers.

$(x - 1) (x + 7) = 0$

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

$x - 1 = 0$

$x + 7 = 0$

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.

$x = 1$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 1) (x + 7) = 0$ true.

$x = 1, - 7$

Solve for x square root of 11-2x = square root of x^2+4x+4

Related Questions

Support LCMGCF.com by disabling your adblocker.