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

1 Answer

Answer:

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#