Finite Math Examples

$b = \sqrt{- 4 + 4 b}$

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.

$\sqrt{- 4 + 4 b} = b$

Step 2

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

${\sqrt{- 4 + 4 b}}^{2} = b^{2}$

Step 3

Simplify each side of the equation.

Tap for more steps...

Step 3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- 4 + 4 b}$ as ${(- 4 + 4 b)}^{\frac{1}{2}}$ .

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

Step 3.2

Simplify the left side.

Tap for more steps...

Step 3.2.1

Simplify ${({(- 4 + 4 b)}^{\frac{1}{2}})}^{2}$ .

Tap for more steps...

Step 3.2.1.1

Multiply the exponents in ${({(- 4 + 4 b)}^{\frac{1}{2}})}^{2}$ .

Tap for more steps...

Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.1.2

Simplify.

$- 4 + 4 b = b^{2}$

Step 4

Solve for $b$ .

Tap for more steps...

Step 4.1

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

$- 4 + 4 b - b^{2} = 0$

Step 4.2

Factor the left side of the equation.

Tap for more steps...

Step 4.2.1

Factor $- 1$ out of $- 4 + 4 b - b^{2}$ .

Tap for more steps...

Step 4.2.1.1

Reorder the expression.

Tap for more steps...

Step 4.2.1.1.1

Move $- 4$ .

$4 b - b^{2} - 4 = 0$

Step 4.2.1.1.2

Reorder $4 b$ and $- b^{2}$ .

$- b^{2} + 4 b - 4 = 0$

Step 4.2.1.2

Factor $- 1$ out of $- b^{2}$ .

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

Step 4.2.1.3

Factor $- 1$ out of $4 b$ .

$- (b^{2}) - (- 4 b) - 4 = 0$

Step 4.2.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$- (b^{2}) - (- 4 b) - 1 \cdot 4 = 0$

Step 4.2.1.5

Factor $- 1$ out of $- (b^{2}) - (- 4 b)$ .

$- (b^{2} - 4 b) - 1 \cdot 4 = 0$

Step 4.2.1.6

Factor $- 1$ out of $- (b^{2} - 4 b) - 1 (4)$ .

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

Step 4.2.2

Factor using the perfect square rule.

Tap for more steps...

Step 4.2.2.1

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

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

Step 4.2.2.2

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

$4 b = 2 \cdot b \cdot 2$

Step 4.2.2.3

Rewrite the polynomial.

$- (b^{2} - 2 \cdot b \cdot 2 + 2^{2}) = 0$

Step 4.2.2.4

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

$- {(b - 2)}^{2} = 0$

Step 4.3

Divide each term in $- {(b - 2)}^{2} = 0$ by $- 1$ and simplify.

Tap for more steps...

Step 4.3.1

Divide each term in $- {(b - 2)}^{2} = 0$ by $- 1$ .

$\frac{- {(b - 2)}^{2}}{- 1} = \frac{0}{- 1}$

Step 4.3.2

Simplify the left side.

Tap for more steps...

Step 4.3.2.1

Dividing two negative values results in a positive value.

$\frac{{(b - 2)}^{2}}{1} = \frac{0}{- 1}$

Step 4.3.2.2

Divide ${(b - 2)}^{2}$ by $1$ .

${(b - 2)}^{2} = \frac{0}{- 1}$

Step 4.3.3

Simplify the right side.

Tap for more steps...

Step 4.3.3.1

Divide $0$ by $- 1$ .

${(b - 2)}^{2} = 0$

Step 4.4

Set the $b - 2$ equal to $0$ .

$b - 2 = 0$

Step 4.5

Add $2$ to both sides of the equation.

$b = 2$