# How do you solve 3x^2+7x+4=0?

Mar 22, 2016

See explanation for a few methods...

#### Explanation:

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Quick Method

Notice that if you invert the sign of the term of odd degree then the sum of the coefficients is zero. That is: $3 - 7 + 4 = 0$

We can deduce that $x = - 1$ is a root and $\left(x + 1\right)$ a factor:

$0 = 3 {x}^{2} + 7 x + 4 = \left(x + 1\right) \left(3 x + 4\right)$

So the other root is $x = - \frac{4}{3}$

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AC Method

Look for a pair of factors of $A C = 3 \cdot 4 = 12$ with sum $B = 7$.

The pair $3 , 4$ works.

Use that pair to split the middle term and factor by grouping:

$0 = 3 {x}^{2} + 7 x + 4$

$= 3 {x}^{2} + 3 x + 4 x + 4$

$= \left(3 {x}^{2} + 3 x\right) + \left(4 x + 4\right)$

$= 3 x \left(x + 1\right) + 4 \left(x + 1\right)$

$= \left(3 x + 4\right) \left(x + 1\right)$

Hence roots $x = - \frac{4}{3}$ and $x = - 1$

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Completing the square

Use the difference of squares identity too:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(6 x + 7\right)$ and $b = 1$

First multiply the equation by ${2}^{2} \cdot 3 = 12$ to cut down on the fractions involved:

$0 = 3 {x}^{2} + 7 x + 4$

becomes

$0 = 36 {x}^{2} + 84 x + 48$

$= {\left(6 x + 7\right)}^{2} - 49 + 48$

$= {\left(6 x + 7\right)}^{2} - {1}^{2}$

$= \left(\left(6 x + 7\right) - 1\right) \left(\left(6 x + 7\right) + 1\right)$

$= \left(6 x + 6\right) \left(6 x + 8\right)$

$= \left(6 \left(x + 1\right)\right) \left(2 \left(3 x + 4\right)\right)$

$= 12 \left(x + 1\right) \left(3 x + 4\right)$

Hence roots $x = - 1$ and $x = - \frac{4}{3}$