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

Aug 15, 2016

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

Explanation:

Factorise to get $\left(3 x - 1\right) \left(x + 2\right) = 0$

hence
$3 x - 1 = 0 \mathmr{and} x + 2 = 0$

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

Aug 15, 2016

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

Explanation:

Find factors of 3 and 2 which subtract to make 5.

We should realise that (3x2) - 1 = 5.
Find the cross products and subtract to get 5.

$3 \text{ "1 " } \Rightarrow 1 \times 1 = 1$
$1 \text{ "2" "rArr 3xx2 = 6 " } \left(6 - 1 = 5\right)$

$3 \text{ "-1 " } \Rightarrow 1 \times - 1 = - 1$
$1 \text{ "+2" "rArr 3xx+2 = +6 " } \left(+ 6 - 1 = + 5\right)$

$\left(3 x - 1\right) \left(x + 2\right) = 0$
One of the factors must be 0.

If $3 x - 1 = 0 , \text{ then } x = \frac{1}{3}$
If $x + 2 = 0 , \text{ then } x = - 2$

Aug 15, 2016

There is only one way to deal with factorization. That is lots of practice.
$x = + \frac{1}{3}$

$x = - 2$

Explanation:

Given:$\text{ } 3 {x}^{2} + 5 x - 2 = 0$
'............................................................................................................

Both 3 and 2 are prime numbers so they can only be the products of:

for 2$\to 1 \times 2$
for 3$\to 1 \times 3$

So without taking any notice of the signs (sorted out afterwards) we have something like: $\textcolor{red}{\left(3 x + 1\right) \left(x + 2\right) \text{ or } \left(3 x + 2\right) \left(x + 1\right)}$
I have kept both $3 x \text{ and } x$ positive.

$\textcolor{red}{\text{Note that the signs in the above will be wrong!}}$

$\textcolor{g r e e n}{\text{Ok. Lets consider the signs by looking at the } + 5 x}$ $\textcolor{g r e e n}{\text{from } 3 {x}^{2} + 5 x - 2}$

Notice that
$\left(+ 3 x\right) \times \left(+ 2\right) = 6 x$
$\left(- 1\right) \times x = - x$

giving $6 x - x = + 5 x$ as required

So the factor form is very likely:

$\left(3 x - 1\right) \left(x + 2\right)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:
Multiplying out the brackets
color(blue)((3x-1))color(brown)((x+2) )" "rarr" " color(brown)(color(blue)(3x)(x+2)color(blue)(" "-1)(x+2))

$\text{ "=3x^2+6x" } - x - 2$

$= 3 {x}^{2} + 5 x - 2$ as required
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we have:

$3 {x}^{2} + 5 x - 2 = 0 \text{ "->" } \left(3 x - 1\right) \left(x + 2\right) = 0$

Note that anything $\times 0 = 0$

This means that the solution is such that

$\left(3 x - 1\right) = 0 \to x = + \frac{1}{3}$

$\left(x + 2\right) = 0 \to x = - 2$