Algebra Examples

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

Step 1

Solve the equation for $y$ .

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Step 1.1

Factor using the perfect square rule.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

Rewrite the polynomial.

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

Step 1.1.5

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

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

Step 1.2

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

$2 y + 1 = 0$

Step 1.3

Solve for $y$ .

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Step 1.3.1

Subtract $1$ from both sides of the equation.

$2 y = - 1$

Step 1.3.2

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

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Step 1.3.2.1

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

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

Step 1.3.2.2

Simplify the left side.

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Step 1.3.2.2.1

Cancel the common factor of $2$ .

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Step 1.3.2.2.1.1

Cancel the common factor.

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

Step 1.3.2.2.1.2

Divide $y$ by $1$ .

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

Step 1.3.2.3

Simplify the right side.

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Step 1.3.2.3.1

Move the negative in front of the fraction.

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

Step 2

Set the numerator equal to zero.

$1 = 0$

Step 3

Since $1 \neq 0$ , there are no solutions.

No solution