SOLUTION: The number of values of a for which [(a^2)-3a+4](x^2) + [(a^2)-5a+6]x + [(a^2)-4]=0 is an identity in x is i) 0 ii) 2 iii) 1 iv) 3

Algebra ->  Square-cubic-other-roots -> SOLUTION: The number of values of a for which [(a^2)-3a+4](x^2) + [(a^2)-5a+6]x + [(a^2)-4]=0 is an identity in x is i) 0 ii) 2 iii) 1 iv) 3      Log On


   

Question 1074872: The number of values of a for which [(a^2)-3a+4](x^2) + [(a^2)-5a+6]x + [(a^2)-4]=0
is an identity in x is
i) 0
ii) 2
iii) 1
iv) 3
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If that is an identities, it should be true for all values of x .

For x=0 , we need to have
a%5E2-4=0 --> a%5E2=4 .
There are values of a that comply with that, namely a=2 and a=-2 ,
but would those values of a work with any and all values of x ?

If a%5E2=4 for x=1 we need to have
%284-3a%2B4%29%2B%284-5a%2B6%29%2B0=0 , which simplifies to
4-3a%2B4%2B4-5a%2B6=0 and further simplifies to
18-8a=0 .
That is not true for either a=2 or a=-2 ,
so there are highlight%280%29 values of a that make that expression an identity.
We could not find even one that would work for x=0 and x=1 ,
so none would work for every x.

ANOTHER WAY:
The only way that ax%5E2%2Bbx%2Bc=0 would be an identity in x
is if a=b=c=0 ,
so the only way that F%28a%29%2Ax%5E2%2BG9a%29%2Ax%2BH%28a%29=0
could be an identity in x for some value of a is if
system%28F%28a%29=0%2CG%28a%29=0%2CH%28a%29=0%29 has some solution.
We could have tried to find the solutions to
a%5E2-3a%2B4=0 (no solution),
a%5E2-5a%2B6=0 (a=3 ans a=2) , and
a%5E2-4=0 (a=2 and a=-2),
and then concluded that no solution worked for all 3 equations.
That may be inefficient, but it may be what was intended,
to make practice solving quadratic equations.
(Of course, if you start by finding that a%5E2-3a%2B4=0 has no real solution,
you do not need to work on solving teh other equations