Solve for j 4j^2+4j+1=0

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

Step 1

Factor using the perfect square rule.

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

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

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

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

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

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

Rewrite the polynomial.

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

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

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

Step 2

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

$2 j + 1 = 0$

Step 3

Solve for $j$ .

Subtract $1$ from both sides of the equation.

$2 j = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $j$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$j = - 0.5$