How do you solve #2x^2-9x-5=0# by factoring?

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Answer 1 · 2015-10-05T10:05:03

The solutions are
#color(blue)(x=-1/2, x=5#

# 2x^2−9x−5=0#

We can Split the Middle Term of this expression to factorise it and thereby 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*-5 = -10#

AND

#N_1 +N_2 = b = -9#

After trying out a few numbers we get #N_1 = -10# and #N_2 =1#

#-10*1 = -10#, and #1+(-10)= -9#

# 2x^2−color(blue)(9x)−5= 2x^2−color(blue)(10x+1x)−5#

#=2x(x-5)+ 1(x−5)#

#color(blue)((2x+1) (x-5)# is the factorised form of the expression.

We now equate the factors to zero to obtain the solutions:

#2x+1=0, color(blue)(x=-1/2#
#x-5=0, color(blue)(x=5#