A function is defined by f(x), value given when f(-2)=9? Hence find f(x)=0?

Question

Find the value of #f(x)=2x^2 +ax-5#
When #f(-2)=9#

This is my answer: #f(-2)=2(-2)^2 +a(-2)-5#
#=8+(-2a)-5#
#=8-2a-5#
#=3-2a#

Therefore:
#9=3-2a#
#=3-(-23)#
#=3+6#
#=9#

Therefore:*
#a=3#

Hence solve #f(x)=0#

I'm a little doubtful here, I believe what I must do is factorise #f(x)#

#f(x)=2x^2+ax-5# --- if #a=3#
#=2x^2+3x-5#
#=2x^2+5x-3x-5#
#=x(2x+5)-(x-1)#
#=(2x+5)(x-1)=0#

What have I done wrong? Any help, notes or correct workings?

Thank you.

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Answer 1 · 2018-05-29T16:31:08

#x=-1, x=5/2#

#f(x)=2x^2 +ax-5#

#f(-2)=9#

#9=2(-2)^2 +a(-2)-5#

#9=8 -2a-5#

#9=3 -2a#

#6=-2a#

#a=-3#

therefore:

#f(x)=2x^2 -3x-5#

#0=2x^2 -3x-5#

#(x + 1) (2 x - 5)=0#

#x=-1, x=5/2#