How do you factor and solve $3x^2+ 2x = 8$ ?

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Answer 1 · 2017-04-29T02:51:05

See a solution process below:

First, subtract $color(red)(8)$ from each side of the equation to put in standard quadratic form:

$3x^2 + 2x - color(red)(8) = 8 - color(red)(8)$

$3x^2 + 2x - 8 = 0$

Next, playing with factors of 3 (1 x 3 and 3 x 1) and factors of 8 (1 x 8, 2 x 4, 4 x 2 and 8 x 1) we can factor this as:

$(3x - 4)(x + 2) = 0$

Now, solve each term for $0$ :

Solution 1)

$3x - 4 = 0$

$3x - 4 + color(red)(4) = 0 + color(red)(4)$

$3x - 0 = 4$

$3x = 4$

$(3x)/color(red)(3) = 4/color(red)(3)$

$(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 4/3$

$x = 4/3$

Solution 2)

$x + 2 = 0$

$x + 2 - color(red)(2) = 0 - color(red)(2)$

$x + 0 = -2$

$x = -2$

The solution is: $x = 4/3$ and $x = -2$

How do you factor and solve 3x^2+ 2x = 83x^2+ 2x = 8?