SOLUTION: If one of the roots of the quadratic equation {{{x^2-(2k-5)x+k^2-5k+56/9=0}}} is twice the other,show that one of the roots is {{{(2k-5)/3}}}.Hence find the values of the constant

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: If one of the roots of the quadratic equation {{{x^2-(2k-5)x+k^2-5k+56/9=0}}} is twice the other,show that one of the roots is {{{(2k-5)/3}}}.Hence find the values of the constant       Log On


   

Question 772758: If one of the roots of the quadratic equation x%5E2-%282k-5%29x%2Bk%5E2-5k%2B56%2F9=0 is twice the other,show that one of the roots is %282k-5%29%2F3.Hence find the values of the constant k.
Thanks!
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
If one of the roots of the quadratic equation x%5E2-%282k-5%29x%2Bk%5E2-5k%2B56%2F9=0 is twice the other,show that one of the roots is %282k-5%29%2F3.
%28%282k-5%29%2F3%29%282%28%282k-5%29%2F3%29%29=0
%282k-5%29%282%282k-5%29%29=0
%282k-5%29%284k-10%29=0
8k%5E2-20k-20k%2B50=0
8k%5E2-40k%2B50=0
k+=+%28-%28-40%29+%2B-+sqrt%28+%28-40%29%5E2-4%2A8%2A50+%29%29%2F%282%2A8%29+
k+=+%2840+%2B-+sqrt%281600-1600+%29%29%2F16+
k+=+%2840+%2B-+sqrt%280+%29%29%2F16+
k+=+40+%2F16+
k+=+5+%2F2+.... the values of the constant k
so, plug it in given equation
x%5E2-%282k-5%29x%2Bk%5E2-5k%2B56%2F9=0
=> x%5E2-%280%29x%2B%285%2F2%29%5E2-5%285%2F2%29%2B56%2F9=0
=> x%5E2%2B25%2F4-25%2F2%2B56%2F9=0
=> x%5E2%2B25%2F4-50%2F4%2B56%2F9=0
=> x%5E2-25%2F4%2B56%2F9=0....-%2825%2A9%29%2F%284%2A9%29%2B%2856%2A4%29%2F%289%2A4%29=-225%2F36%2B216%2F36=-1%2F36
=> x%5E2-1%2F36=0
roots:
%282k-5%29%2F3=>%282%285+%2F2%29-5%29%2F3=%285-5%29%2F3=0
2%282k-5%29%2F3=>2%282%285+%2F2%29-5%29%2F3=2%285-5%29%2F3=0
+graph%28600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2-1%2F36%29+

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
x² - (2k-5)x + (k²-5k+56/9) = 0

Use these facts of quadratics with leading coefficient 1:
For the quadratic equation x²+Ax+B=0
The sum of the roots  is -A
The product of the roots is B

Let r = the root that is NOT twice the other

Then the root which IS twice the other is 2r

Sum of roots = r+2r = 3r

Therefore 

         3r = 2k-5

and       r = %282k-5%29%2F3

That was the first thing you were to show.

Product of roots = (r)(2r) = 2r²

Therefore 

        2r² = k²-5k+56%2F9

So we have the system of equations:

1st:     3r = 2k-5
2nd:    2r² = k²-5k+56%2F9

Clear the 2nd equation of fractions:

       18r² = 9k²-45k+56

Square both sides of the 1st

        9r² = 4k²-20k+25

Multiply through by 2

       18r² = 8k²-40k+50

Set the expressions for 18x² equal:

 9k²-45k+56 = 8k²-40k+50

k² - 5k + 6 = 0
 (k-3)(k-2) = 0

k-3=0;  k-2=0
  k=3;    k=2

So those are the 2 values of k, which is what was asked for.

-------------------------------------------------

We weren't asked for the roots, but we could find the roots of the given
 equation very easily:

for k = 3, substitute in

3r = 2k-5
3r = 2(3)-5
3r = 6-5
3r = 1
 r = 1%2F3

So for k = 3, one root is x = 1%2F3 and the other is x = 2%2F3 

for k = 2, substitute in

3r = 2k-5
3r = 2(2)-5
3r = 4-5
3r = -1
 r = -1%2F3

So for k = 2, one root is x = -1%2F3 and the other is x = -2%2F3

Edwin