How do you solve $(3 x + 3) (x + 2) = 2$ $(3x+3)(x+2)=2$ ?

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How do you solve $(3 x + 3) (x + 2) = 2$ $(3x+3)(x+2)=2$ ?

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Answer 1 · 2015-07-10T07:33:01

$x = \frac{- 9 + \sqrt{33}}{6} or \frac{- 9 - \sqrt{33}}{6}$ $x = (-9 + sqrt(33)) / 6 or (-9 - sqrt(33)) / 6$

Explanation:

Simplify the right-hand-side by expanding,

$(3 x + 3) (x + 2) = 3 x^{2} + 9 x + 6$ $(3x+3)(x+2) = 3x^2 + 9x + 6$

Bring the 2 to the left hand side [Think of subtracting both sides by 2]. Then the equation becomes:

$3 x^{2} + 9 x + 4 = 0$ $3x^2 + 9x + 4 = 0$

Solve this using the quadratic formula:

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 12}}{6} = \frac{- 9 \pm \sqrt{33}}{6}$ $x = (-9 +- sqrt(81 - 4*12))/6 = (-9 +- sqrt(33)) / 6$

Thus, $x = {\frac{- 9 + \sqrt{33}}{6}, \frac{- 9 - \sqrt{33}}{6}}$ $x = { (-9 + sqrt(33)) / 6, (-9 - sqrt(33)) / 6 }$

Answer 2 · 2015-07-10T07:37:09

$x = \frac{- 9 + \sqrt{33}}{6},$ $x=(-9+sqrt(33))/6,$ $\frac{- 9 - \sqrt{33}}{6}$ $(-9-sqrt(33))/6$

Explanation:

$(3 x + 3) (x + 2) = 2$ $(3x+3)(x+2)=2$

FOIL the left side of the equation.

$3 x^{2} + 6 x + 3 x + 6 = 2$ $3x^2+6x+3x+6=2$ =

$3 x^{2} + 9 x + 6 = 2$ $3x^2+9x+6=2$

Subtract $2$ $2$ from both sides.

$3 x^{2} + 9 x + 6 - 2 = 0$ $3x^2+9x+6-2=0$ =

$3 x^{2} + 9 x + 4 = 0$ $3x^2+9x+4=0$

The equation is now in the form of a quadratic equation: $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , where $a = 3;$ $a=3;$ $b = 9;$ $b=9;$ and $c = 4$ $c=4$ .

Use the quadratic formula to solve this equation.

Quadratic Formula

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ =

$x = \frac{- 9 \pm \sqrt{9^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3}$ $x=(-9+-sqrt(9^2-4*3*4))/(2*3)$ =

$x = \frac{- 9 \pm \sqrt{81 - 48}}{6}$ $x=(-9+-sqrt(81-48))/6$ =

$x = \frac{- 9 \pm \sqrt{33}}{6}$ $x=(-9+-sqrt(33))/6$

Two answers for $x$ $x$ .

$x = \frac{- 9 + \sqrt{33}}{6}$ $x=(-9+sqrt(33))/6$

$x = \frac{- 9 - \sqrt{33}}{6}$ $x=(-9-sqrt(33))/6$

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How do you solve $(3 x + 3) (x + 2) = 2$ $(3x+3)(x+2)=2$ ?

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