If x-3 and x-1/3 are both factors of ax^2+5x+b ; show that a=b
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if x-3 is a factor, then ax²+5x+b=0 when x=3
=>9a+15+b=0-------(1)
if x-1/3 is a factor then ax²+5x+b=0 when x=1/3
=>a/9+5/3+b=0------(2)
Now,
(1)-(2)=>a(9-1/9)+15-5/3=0
=>80a/9=-40/3
=>a=-3/2
So, (2)=>-1/6 +5/3 +b =0
=>b=-3/2
So, a=b [showed]
=>9a+15+b=0-------(1)
if x-1/3 is a factor then ax²+5x+b=0 when x=1/3
=>a/9+5/3+b=0------(2)
Now,
(1)-(2)=>a(9-1/9)+15-5/3=0
=>80a/9=-40/3
=>a=-3/2
So, (2)=>-1/6 +5/3 +b =0
=>b=-3/2
So, a=b [showed]
emma18:
Thanks
Hey! Quick question! How did you get -1/6?
I just plugged the value of a in equation (2) and accomplished a simple arithmetic to get it.
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Hey
Here is your answer
Hope this helps you
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