Question

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

Answer 1

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 `(x+1)` a factor:

`0 = 3x^2+7x+4=(x+1)(3x+4)`

So the other root is `x = -4/3`

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AC Method

Look for a pair of factors of `AC = 3*4 = 12` with sum `B=7`.

The pair `3, 4` works.

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

`0 = 3x^2+7x+4`

`=3x^2+3x+4x+4`

`=(3x^2+3x)+(4x+4)`

`=3x(x+1)+4(x+1)`

`=(3x+4)(x+1)`

Hence roots `x=-4/3` and `x = -1`

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Completing the square

Use the difference of squares identity too:

`a^2-b^2 = (a-b)(a+b)`

with `a = (6x+7)` and `b=1`

First multiply the equation by `2^2*3 = 12` to cut down on the fractions involved:

`0 = 3x^2+7x+4`

becomes

`0 = 36x^2+84x+48`

`=(6x+7)^2-49+48`

`=(6x+7)^2-1^2`

`=((6x+7)-1)((6x+7)+1)`

`=(6x+6)(6x+8)`

`=(6(x+1))(2(3x+4))`

`=12(x+1)(3x+4)`

Hence roots `x=-1` and `x=-4/3`