SOLUTION: find k so thqqt the equation x2-11x+k=0 andx2-14x+2k=0 may have a common root

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Question 551052: find k so thqqt the equation x2-11x+k=0 andx2-14x+2k=0 may have a common root

Found 2 solutions by oberobic, KMST:
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
f(x) = x^2 -11x + k
g(x) = x^2 -14x + 2k
.
x = 0 defines the roots
But that reveals we would be looking for a point where k = 2k.
That exists only at k=0.
Given k=0, we can redefine f(x) and g(x):
f(x) = x*(x-11)
g(x) = x*(x-14)
They share the common root 0.
.
+graph%28500%2C500%2C-5%2C20%2C-40%2C5%2Cx%5E2-11%2Ax%2B0%2Cx%5E2-14%2Ax%2B2%2A0%29+

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
An obvious solution is k=0.
However, k=24 is another solution, giving both equations,
x%5E2-11x%2B24=0 and x%5E2-14x%2B48=0
the common root x=8.
graph%28300%2C300%2C-2%2C13%2C-8%2C32%2Cx%5E2-11x%2B24%2Cx%5E2-14x%2B48%29
A common root would be a solution of the non-linear system of non-linear equations:
x%5E2-11x%2Bk=0
x%5E2-14x%2B2k=0
We can combine them to give
x%5E2-11x%2Bk=x%5E2-14x%2B2k --> 3x=k --> x=k%2F3
which we could use, along with the friendliest of the quadratic equations above to search for all possible solutions.
x%5E2-14x%2B2k=0 has the solutions
x=7%2B-sqrt%2849-2k%29
Combining them, we get
k%2F3=7%2B-sqrt%2849-2k%29 --> k=21+%2B-+3sqrt%2849-2k%29 --> %28k-21%29%5E2=3%5E2%2849-2k%29 --> k%5E2-42k%2B441=441-18x --> k%5E2-24k=0 --> k%28k-24%29=0
So we get two answers k=0 and k=24, and both verify.
For k=0, we get x=0/3=0.
For k=24 we get x=24/3=8.
The solutions are the (x,k) pairs (0,0) and (8,24).