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

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Answer 1 · 2015-11-26T14:56:43

The solutions are:

$color(blue)(x=-1/2 color(white)(...........)$

$color(blue)(x=-2$

$2x^2+5x+2=0$

We can Split the Middle Term of this expression to factorise it and then find the solutions.

In this technique, if we have to factorise an expression like $ax^2 + bx + c$ , we need to think of 2 numbers such that:

$N_1*N_2 = a*c = 2*2 =4$
AND
$N_1 +N_2 = b = 5$

After trying out a few numbers we get $N_1 = 4$ and $N_2 =1$
$4*1 =4$ , and $4+1= 5$

$2x^2+color(blue)(5x)+2=2x^2+color(blue)(4x+1x)+2=0$

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

$color(blue)((2x+1)(x+2)=0$

Now we equate the factors to zero:

$2x+1=0, color(blue)(x=-1/2$

$x+2=0, color(blue)(x=-2$

How do you solve 2x^2+5x+2=02x^2+5x+2=0?