Find the Discriminant 4y^2+4y+1=0

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

Step 1

Solve the equation for $y$ .

Factor using the perfect square rule.

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

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

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

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

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

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

Rewrite the polynomial.

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

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

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

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

$2 y + 1 = 0$

Solve for $y$ .

Subtract $1$ from both sides of the equation.

$2 y = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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