How do I solve this quadratic equation?

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How do I solve this quadratic equation?

$6 x^{2} + 7 x + 2 = 0$ $6x^2 + 7x +2=0$

3 Answers

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Answer 1 · 2018-03-09T00:39:00

$x = - \frac{1}{2}$ $x = -1/2$ and $x = - \frac{2}{3}$ $x = -2/3$

Explanation:

$6 x^{2} + 7 x + 2$ $6x^2 + 7x + 2$

can be factored into a binomial,

$(3 x + \frac{3}{2}) (2 x + \frac{4}{3})$ $(3x+3/2)(2x+4/3)$

By setting a factor to zero we can solve for an x value
$3 x + \frac{3}{2} = 0$ $3x+3/2 = 0$
$x = - \frac{1}{2}$ $x = -1/2$

$2 x + \frac{4}{3} = 0$ $2x+4/3 = 0$
$x = - \frac{2}{3}$ $x= -2/3$

Answer 2 · 2018-03-09T00:41:26

$x = - \frac{1}{2}, - \frac{2}{3}$ $x=-1/2, -2/3$

Explanation:

We can solve this quadratic with the strategy factoring by grouping. Here, we will rewrite the $x$ $x$ term as the sum of two terms, so we can split them up and factor. Here's what I mean:

$6 x^{2} + 7 x + 2 = 0$ $6x^2+color(blue)(7x)+2=0$

This is equivalent to the following:

$6 x^{2} + 3 x + 4 x + 2 = 0$ $6x^2+color(blue)(3x+4x)+2=0$

Notice, I only rewrote $7 x$ $7x$ as the sum of $3 x$ $3x$ and $4 x$ $4x$ so we can factor. You'll see why this is useful:

$6 x^{2} + 3 x + 4 x + 2 = 0$ $color(red)(6x^2+3x)+color(orange)(4x+2)=0$

We can factor a $3 x$ $3x$ out of the red expression, and a $2$ $2$ out of the orange expression. We get:

$3 x (2 x + 1) + 2 (2 x + 1) = 0$ $color(red)(3x(2x+1))+color(orange)(2(2x+1))=0$

Since $3 x$ $3x$ and $2$ $2$ are being multiplied by the same term ( $2 x + 1$ $2x+1$ ), we can rewrite this equation as:

$(3 x + 2) (2 x + 1) = 0$ $(3x+2)(2x+1)=0$

We now set both factors equal to zero to get:

$3 x + 2 = 0$ $3x+2=0$

$\Rightarrow 3 x = - 2$ $=>3x=-2$

$\Rightarrow x = - \frac{2}{3}$ $color(blue)(=>x=-2/3)$

$2 x + 1 = 0$ $2x+1=0$

$\Rightarrow 2 x = - 1$ $=>2x=-1$

$\Rightarrow x = - \frac{1}{2}$ $color(blue)(=>x=-1/2)$

Our factors are in blue. Hope this helps!

Answer 3 · 2018-03-09T01:06:25

$- \frac{1}{2} = x = - \frac{2}{3}$ $-1/2=x=-2/3$

Explanation:

Hmm...
We have:
$6 x^{2} + 7 x + 2 = 0$ $6x^2+7x+2=0$ Since $x^{2}$ $x^2$ is being multiplied by a number here, let's multiply $a$ $a$ and $c$ $c$ in $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$a \cdot c = 6 \cdot 2 \Rightarrow 12$ $a*c=6*2=>12$

We ask ourselves: Do any of the factors of $12$ $12$ add up to $7$ $7$ ?

Let's see...

$1 \cdot 12$ $1*12$ Nope.

$2 \cdot 6$ $2*6$ Nope.

$3 \cdot 4$ $3*4$ Yep.

We now rewrite the equation like the following:

$6 x^{2} + 3 x + 4 x + 2 = 0$ $6x^2+3x+4x+2=0$ (The order of $3 x$ $3x$ and $4 x$ $4x$ does not matter.)

Let's separate the terms like this:

$(6 x^{2} + 3 x) + (4 x + 2) = 0$ $(6x^2+3x)+(4x+2)=0$ Factor each parenthesis.

$\Rightarrow 3 x (2 x + 1) + 2 (2 x + 1) = 0$ $=>3x(2x+1)+2(2x+1)=0$

For better understanding, we let $n = 2 x + 1$ $n=2x+1$

Replace $2 x + 1$ $2x+1$ with $n$ $n$ .

$\Rightarrow 3 x n + 2 n = 0$ $=>3xn+2n=0$ Now, we see that each group have $n$ $n$ in common.

Let's factor each term.

$\Rightarrow n (3 x + 2) = 0$ $=>n(3x+2)=0$ Replace $n$ $n$ with $2 x + 1$ $2x+1$

$\Rightarrow (2 x + 1) (3 x + 2) = 0$ $=>(2x+1)(3x+2)=0$

Either $2 x + 1 = 0$ $2x+1=0$ or $3 x + 2 = 0$ $3x+2=0$

Let's solve each case.

$2 x + 1 = 0$ $2x+1=0$

$2 x = - 1$ $2x=-1$

$x = - \frac{1}{2}$ $x=-1/2$ That's one answer.

$3 x + 2 = 0$ $3x+2=0$

$3 x = - 2$ $3x=-2$

$x = - \frac{2}{3}$ $x=-2/3$ That's another.

Those two are our answers!

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How do I solve this quadratic equation?

$6 x^{2} + 7 x + 2 = 0$ $6x^2 + 7x +2=0$

3 Answers

Explanation:

Explanation:

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$6 x^{2} + 7 x + 2 = 0$ $6x^2 + 7x +2=0$