Solve the Rational Equation for x square root of (x-1)^2=1-x

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

Divide $2$ by $2$ .

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

Simplify the left side.

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

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

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

Multiply $2$ by $1$ .

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

Simplify the right side.

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

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

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

Multiply $- x$ by $1$ .

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

Multiply $- 1$ by $1$ .

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

Rewrite using the commutative property of multiplication.

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

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Subtract $x$ from $- x$ .

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

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.

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

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

Rewrite.

$1 - 2 x + x^{2} = 0 + 0 + {(x - 1)}^{2}$

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $- 1$ to the left of $x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Rewrite $- 1 x$ as $- x$ .

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

Multiply $- 1$ by $- 1$ .

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

Subtract $x$ from $- x$ .

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

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

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

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

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

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

Combine the opposite terms in $1 - 2 x + x^{2} - x^{2} + 2 x$ .

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

$1 - 2 x + 0 + 2 x = 1$

Add $1 - 2 x$ and $0$ .

$1 - 2 x + 2 x = 1$

Add $- 2 x$ and $2 x$ .

$1 + 0 = 1$

Add $1$ and $0$ .

$1 = 1$

Since $1 = 1$ , the equation will always be true.

Always true

Step 4

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$

Solve the Rational Equation for x square root of (x-1)^2=1-x

Related Questions

Support LCMGCF.com by disabling your adblocker.