SOLUTION: if on root of the equation 2x^2-kx+64=0 is twice the other root, then find the value of k

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Question 763902: if on root of the equation 2x^2-kx+64=0 is twice the other root, then find the value of k
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let the roots be a and a%2F2.
If we knew the coefficients and a was a nice round number we may solve such an equation by factoring.
Even when the roots are ugly irrational numbers, a polynomial can be written in factored form if you know the zeros.
In this case, the factored equation would be
2%28x-a%2F2%29%28x-a%29=0 or %282x-a%29%28x-a%29=0
Re-multiplying, we get
2x%5E2-3ax%2Ba%5E2=0
If that equation is equivalent to 2x%5E2-kx%2B64=0 (both with 2 for a leading coefficient),
then the other copefficientsa are also the same, so
a%5E2=64 and -3a+=+-k-->3a=k
system%28a%5E2=64%2C3a=k%29 --> system%28a=8%2Cor%2Ca=-8%29 --> system%28k=24%2Cor%2Ck=-24%29