Question

p(x)=ax²+(a-15)x+1 and p(x)=2x²-3x+1/b have real coefficients and the same roots. What's the value of a + b?

a)12
b)9
c)3
d)6

How can I start solving this?

Answer 1

Okay, if they have the same roots, the discriminant, which equals:
b^2-4ac, is 0, so start off with:
b^2-4ac=0

Answer 2

oh wait nevermind that haha.. but still find the discriminant

Answer 3

i just saw that u had two equations, i misread the question..

Answer 4

oops, second one is q(x)=2x²-3x+1/b

Answer 5

Okay, what is x equal to?
x1 = (-b - sqrt(b^2-4ac))/(2a)
x2 = (-b + sqrt(b^2-4ac))/(2a)
right?

Answer 6

okay let's just look at x1 for now. x1 from the first equation must equal to x1 from the 2nd one right.. so:
(-b - sqrt(b^2-4ac))/(2a) = (-b - sqrt(b^2-4ac))/(2a)
and if we rewrite this in terms of things given..

Answer 7

1) First equation: a = a; b = a-15; c = 1
2) Second equation: a = 2; b = -3; c = 1/b

Answer 8

Plug those all into the equation that i wrote above:
( -(a-15) - sqrt[(a-15)^2-4*a*1] )/(2a) = ( 3 - sqrt[9 - 4 * 2 * (1/b)] )/4

Answer 9

Simplify the mess a little:
( -a + 15 - sqrt[ (a-15)^2 - 4a] )/(2a) = ( 3 - sqrt[9-8/b] )/4

Answer 10

oh god.. now we gotta do the same thing for x2..

Answer 11

Start with
p(x)=ax²+(a-15)x+1 and p(x)=2x²-3x+1/b have real coefficients
and the same roots.

The same roots means
ax²+(a-15)x+1 =0
2x²-3x+1/b = 0

for the 2nd equation, multiply both sides (all terms) by b to get
2b x^2 -3b +1 =0
and of course, ax²+(a-15)x+1 =0

now match coefficients of each term:
a= 2b
(a-15)= -3b

solve for a and b. One way: sub in 2b for a in the 2nd equation
2b-15= -3b
5b=15
b=3
a=6

(a+b) = 9

Answer 12

( -(a-15) + sqrt[(a-15)^2-4*a*1] )/(2a) = ( 3 + sqrt[9 - 4 * 2 * (1/b)] )/4

Answer 13

omg phi.... lol im doing this whole freakin thing like an idiot.. fml

Answer 14

Phi's right, all u gotta do is just match the coefficients..

Answer 15

* 2b x^2 -3b x +1 =0 (in case it wasn't obvious)

Answer 16

thank you guys very much! haha, i told ya bahrom, you like the hardest ways :P

Answer 17

lol melinda.. that got me into trouble soo many times... but i always find the easy way at the end, like 5 mins before the exam's over