Finite Math Examples

$4 r^{2} + 20 r + 25 = 0$

Step 1

Factor using the perfect square rule.

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Step 1.1

Rewrite $4 r^{2}$ as ${(2 r)}^{2}$ .

${(2 r)}^{2} + 20 r + 25 = 0$

Step 1.2

Rewrite $25$ as $5^{2}$ .

${(2 r)}^{2} + 20 r + 5^{2} = 0$

Step 1.3

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

$20 r = 2 \cdot (2 r) \cdot 5$

Step 1.4

Rewrite the polynomial.

${(2 r)}^{2} + 2 \cdot (2 r) \cdot 5 + 5^{2} = 0$

Step 1.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 r$ and $b = 5$ .

${(2 r + 5)}^{2} = 0$

Step 2

Set the $2 r + 5$ equal to $0$ .

$2 r + 5 = 0$

Step 3

Solve for $r$ .

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Step 3.1

Subtract $5$ from both sides of the equation.

$2 r = - 5$

Step 3.2

Divide each term in $2 r = - 5$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 r = - 5$ by $2$ .

$\frac{2 r}{2} = \frac{- 5}{2}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 r}{2} = \frac{- 5}{2}$

Step 3.2.2.1.2

Divide $r$ by $1$ .

$r = \frac{- 5}{2}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Move the negative in front of the fraction.

$r = - \frac{5}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$r = - \frac{5}{2}$

Decimal Form:

$r = - 2.5$

Mixed Number Form:

$r = - 2 \frac{1}{2}$