Solve for x 5+2x = square root of x^2-2x+1

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

Step 1

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

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

Step 2

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

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

Step 3

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Rewrite ${(5 + 2 x)}^{2}$ as $(5 + 2 x) (5 + 2 x)$ .

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

Expand $(5 + 2 x) (5 + 2 x)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$x^{2} - 2 x + 1 = 25 + 5 (2 x) + 2 x \cdot 5 + 2 x (2 x)$

Multiply $2$ by $5$ .

$x^{2} - 2 x + 1 = 25 + 10 x + 2 x \cdot 5 + 2 x (2 x)$

Multiply $5$ by $2$ .

$x^{2} - 2 x + 1 = 25 + 10 x + 10 x + 2 x (2 x)$

Rewrite using the commutative property of multiplication.

$x^{2} - 2 x + 1 = 25 + 10 x + 10 x + 2 \cdot 2 x \cdot x$

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

Move $x$ .

$x^{2} - 2 x + 1 = 25 + 10 x + 10 x + 2 \cdot 2 (x \cdot x)$

Multiply $x$ by $x$ .

$x^{2} - 2 x + 1 = 25 + 10 x + 10 x + 2 \cdot 2 x^{2}$

Multiply $2$ by $2$ .

$x^{2} - 2 x + 1 = 25 + 10 x + 10 x + 4 x^{2}$

Add $10 x$ and $10 x$ .

$x^{2} - 2 x + 1 = 25 + 20 x + 4 x^{2}$

Step 4

Solve for $x$ .

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

Subtract $20 x$ from both sides of the equation.

$x^{2} - 2 x + 1 - 20 x = 25 + 4 x^{2}$

Subtract $4 x^{2}$ from both sides of the equation.

$x^{2} - 2 x + 1 - 20 x - 4 x^{2} = 25$

Subtract $4 x^{2}$ from $x^{2}$ .

$- 3 x^{2} - 2 x + 1 - 20 x = 25$

Subtract $20 x$ from $- 2 x$ .

$- 3 x^{2} - 22 x + 1 = 25$

Subtract $25$ from both sides of the equation.

$- 3 x^{2} - 22 x + 1 - 25 = 0$

Subtract $25$ from $1$ .

$- 3 x^{2} - 22 x - 24 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 3 x^{2} - 22 x - 24$ .

Factor $- 1$ out of $- 3 x^{2}$ .

$- (3 x^{2}) - 22 x - 24 = 0$

Factor $- 1$ out of $- 22 x$ .

$- (3 x^{2}) - (22 x) - 24 = 0$

Rewrite $- 24$ as $- 1 (24)$ .

$- (3 x^{2}) - (22 x) - 1 \cdot 24 = 0$

Factor $- 1$ out of $- (3 x^{2}) - (22 x)$ .

$- (3 x^{2} + 22 x) - 1 \cdot 24 = 0$

Factor $- 1$ out of $- (3 x^{2} + 22 x) - 1 (24)$ .

$- (3 x^{2} + 22 x + 24) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 24 = 72$ and whose sum is $b = 22$ .

Factor $22$ out of $22 x$ .

$- (3 x^{2} + 22 (x) + 24) = 0$

Rewrite $22$ as $4$ plus $18$

$- (3 x^{2} + (4 + 18) x + 24) = 0$

Apply the distributive property.

$- (3 x^{2} + 4 x + 18 x + 24) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((3 x^{2} + 4 x) + 18 x + 24) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (x (3 x + 4) + 6 (3 x + 4)) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 4$ .

$- ((3 x + 4) (x + 6)) = 0$

Remove unnecessary parentheses.

$- (3 x + 4) (x + 6) = 0$

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

$3 x + 4 = 0$

$x + 6 = 0$

Set $3 x + 4$ equal to $0$ and solve for $x$ .

Set $3 x + 4$ equal to $0$ .

$3 x + 4 = 0$

Solve $3 x + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$3 x = - 4$

Divide each term in $3 x = - 4$ by $3$ and simplify.

Divide each term in $3 x = - 4$ by $3$ .

$\frac{3 x}{3} = \frac{- 4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 4}{3}$

Divide $x$ by $1$ .

$x = \frac{- 4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{4}{3}$

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

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

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $- (3 x + 4) (x + 6) = 0$ true.

$x = - \frac{4}{3}, - 6$

Step 5

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

$x = - \frac{4}{3}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{4}{3}$

Decimal Form:

$x = - 1.\overline{3}$

Mixed Number Form:

$x = - 1 \frac{1}{3}$

Solve for x 5+2x = square root of x^2-2x+1

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