Finite Math Examples

$4 t^{2} + 4 t + 1 = 0$

Step 1

Factor using the perfect square rule.

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Step 1.1

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

${(2 t)}^{2} + 4 t + 1 = 0$

Step 1.2

Rewrite $1$ as $1^{2}$ .

${(2 t)}^{2} + 4 t + 1^{2} = 0$

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.

$4 t = 2 \cdot (2 t) \cdot 1$

Step 1.4

Rewrite the polynomial.

${(2 t)}^{2} + 2 \cdot (2 t) \cdot 1 + 1^{2} = 0$

Step 1.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 t$ and $b = 1$ .

${(2 t + 1)}^{2} = 0$

Step 2

Set the $2 t + 1$ equal to $0$ .

$2 t + 1 = 0$

Step 3

Solve for $t$ .

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Step 3.1

Subtract $1$ from both sides of the equation.

$2 t = - 1$

Step 3.2

Divide each term in $2 t = - 1$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 t = - 1$ by $2$ .

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

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.

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

Step 3.2.2.1.2

Divide $t$ by $1$ .

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

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.

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