does anyone know the solution to 3x2013x-10=0

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- Sep 5th 2007, 02:23 PM #1lynnGuest

- Sep 5th 2007, 02:34 PM #2
Is this supposed to be

$\displaystyle 3x^2 - 13x - 10 = 0$?

When all else fails, you can always resort to the quadratic equation. However since this factors, I'll show you how to do it that way.

This is called the "ac" method:

Multiply the coefficient of the $\displaystyle x^2$ term and the constant term:

$\displaystyle 3 \cdot -10 = -30$

Now list all the pairs of factors of -30:

1, -30

2, -15

3, -10

5, -6

6, -5

10, -3

15, -2

30, -1

Now add each pair of factors:

1 + (-30) = -29

2 + (-15) = -13

3 + (-10) = -7

etc.

If the coefficient of the x term appears in this list then the quadratic factors. In this case we are looking for -13, and we have that 2 + (-15) = -13.

So split the linear term into two pieces using the sum as your guide:

$\displaystyle 3x^2 - 13x - 10 = 0$

$\displaystyle 3x^2 + (2x - 15x) - 10 = 0$

Now regroup:

$\displaystyle (3x^2 + 2x) + (-15x - 10) = 0$

Now factor each term:

$\displaystyle x(3x + 2) - 5(3x + 2) = 0$

Now you have two terms with a common factor, the 3x + 2. So factor again:

$\displaystyle (x - 5)(3x + 2) = 0$

Now use the fact that if you have two real numbers a and b and ab = 0, then we must have that a = 0 or b = 0:

$\displaystyle x - 5 = 0 \implies x = 5$

or

$\displaystyle 3x + 2 = 0 \implies x = -\frac{2}{3}$

So your two solutions are $\displaystyle x = 5, -\frac{3}{2}$.

-Dan