Question

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

Answer 1

The solutions are:

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

`color(blue)(x=-2`

Explanation:

`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`