How do you solve $3 x^{2} + 5 x + 2 = 0$ $3x^2+5x+2=0$ ?

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Answer 1 · 2016-03-13T21:20:13

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

3 and 2 are both prime. Thus the only factors of them is 1 and 3 for 3; 1 and 2 for 2.

Assuming there are integer solutions for $x$ $x$ then it has to be true that one of the following is true. If not, then it is down to the formula.

$(3 x \pm 2) (x \pm 1)$ $(3x+-2)(x+-1)$ as the first option
$(3 x \pm 1) (x \pm 2)$ $(3x+-1)(x+-2)$ as the second option

In the given question:
The constant 2 is positive so the signs in the brackets are the same.
The $5 x$ $5x$ is positive so both signs are +

$Test 1$ $color(blue)("Test 1")$

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

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

$= 3 x^{2} + 5 x + 2$ $color(blue)( =" " 3x^2+5x+2)$

How do you solve 3x^2+5x+2=0 3x2+5x+2=03x^2+5x+2=0 ?