Solve by Factoring 49x^2+42x+9=0

$49 x^{2} + 42 x + 9 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $49 x^{2}$ as ${(7 x)}^{2}$ .

${(7 x)}^{2} + 42 x + 9 = 0$

Rewrite $9$ as $3^{2}$ .

${(7 x)}^{2} + 42 x + 3^{2} = 0$

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

$42 x = 2 \cdot (7 x) \cdot 3$

Rewrite the polynomial.

${(7 x)}^{2} + 2 \cdot (7 x) \cdot 3 + 3^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 7 x$ and $b = 3$ .

${(7 x + 3)}^{2} = 0$

Step 2

Set the $7 x + 3$ equal to $0$ .

$7 x + 3 = 0$

Step 3

Solve for $x$ .

Subtract $3$ from both sides of the equation.

$7 x = - 3$

Divide each term in $7 x = - 3$ by $7$ and simplify.

Divide each term in $7 x = - 3$ by $7$ .

$\frac{7 x}{7} = \frac{- 3}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 3}{7}$

Divide $x$ by $1$ .

$x = \frac{- 3}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{7}$