SOLUTION: Please help me with these questions : 1) The equations ax^2 +bx+c =0 and bx^2 +ax+c=0 have the same roots. Show that a+b+c=0 where a is not =b and c is not =0. 2) show that th

Algebra ->  Radicals -> SOLUTION: Please help me with these questions : 1) The equations ax^2 +bx+c =0 and bx^2 +ax+c=0 have the same roots. Show that a+b+c=0 where a is not =b and c is not =0. 2) show that th      Log On


   



Question 824722: Please help me with these questions :
1) The equations ax^2 +bx+c =0 and bx^2 +ax+c=0 have the same roots. Show that a+b+c=0 where a is not =b and c is not =0.
2) show that the equation (x-k)^ 2 +2x-k has real roots and hence show that 1

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1) If the equations ax%5E2%2Bbx%2Bc=0 and bx%5E2%2Bax%2Bc=0 ,
where a%3C%3Eb and c%3C%3E0 , have a common root, that root will satisfy
ax%5E2%2Bbx%2Bc=bx%5E2%2Bax%2Bc .

Solving:
ax%5E2%2Bbx%2Bc=bx%5E2%2Bax%2Bc
ax%5E2%2Bbx%2Bcross%28c%29=bx%5E2%2Bax%2Bcross%28c%29
ax%5E2%2Bbx=bx%5E2%2Bax
ax%5E2-bx%5E2%2Bbx-ax=0
x%28ax-bx%2Bb-a%29=0
If x=0 were a solution of ax%5E2%2Bbx%2Bc=0 , c would be zero:
a%2A0%5E2%2Bb%2A0%2Bc=0%2B0%2Bc=c=0 .
So we know that,
system%28x%3C%3E0%2Cx%28ax-bx%2Bb-a%29=0%29 --> %28ax-bx%2Bb-a%29=0
%28ax-bx%2Bb-a%29=0
%28a-b%29x%2Bb-a=0
%28a-b%29x=a-b
We can divide both sides of the equal sign by %28a-b%29%3C%3E0 , to get
highlight%28x=1%29.
(We know that %28a-b%29%3C%3E0 because a%3C%3Eb ).

Since we have concluded that only x=1 could be a solution to both equations,
substituting 1 for x must make both equations true.
Substituting into ax%5E2%2Bbx%2Bc=0 ,
we get
0=ax%5E2%2Bbx%2Bc=a%2A1%5E2%2Bb%2A1%2Bc
0=ax%5E2%2Bbx%2Bc=a%2Bb%2Bc
highlight%28a%2Bb%2Bc=0%29
We get the same using the other equation.
The two equations could be x%5E2-3x%2B2=0 , with roots x=1 and x=2 ,
and -3x%5E2%2Bx%2B2=0 , with roots x=1 and x=-2%2F3.

NOTE:
The only solution that we found that could be a common solution to both equations was x=1 .
There must be two more, different solutions, one for each equation.
It cannot be x=1 as a double solution to both equations,
because that would require a=b .
If x=1 were the only solution to both equations, the equations would be
0=ax%5E2%2Bbx%2Bc=a%28x-1%29%5E2=a%28x%5E2-2x%2B1%29=ax%5E2-2ax%2Bhighlight%28a%29 , and
0=bx%5E2%2Bax%2Bc=b%28x-1%29%5E2=b%28x%5E2-2x%2B1%29=bx%5E2-2bx%2Bhighlight%28b%29 .
The independent term in both should be
c=highlight%28a%29 and c=highlight%28b%29 .
That would require a=b .

2) %28x-k%29%5E+2+%2B2x-k=0
%28x-k%29%5E+2+%2B2x-k-k=-k
%28x-k%29%5E+2+%2B2x-2k=-k
%28x-k%29%5E+2+%2B2%28x-k%29=-k
%28x-k%29%5E+2+%2B2%28x-k%29%2B1=1-k
%28%28x-k%29+%2B1%29%5E+2=1-k
So 1-k%3E=0 --> 1%3E=k <--> k%3C=1

Alternatively,
%28x-k%29%5E+2+%2B2x-k=0
x%5E2-2kx%2Bk%5E2%2B2x-k=0
x%5E2%2B%282-2k%29x%2Bk%5E2-k=0
For real solutions we need the discriminant to be non negative
%282-2k%29%5E2-4%2A1%2A%28k%5E2-k%29%3E=0
4-8k%2B4k%5E2-4k%5E2%2B4k%3E=0
4-8k%2B4k%3E=0
4%3E=4k
1%3E=k <--> k%3C=1