SOLUTION: Please help me solve this problem : Find the value of the constant k such that the ratio of the two solution of the quadratic equation 2x2 - kx+ k +2 = 0 is 3:2

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Question 1075689: Please help me solve this problem : Find the value of the constant k such that the ratio of the two
solution of the quadratic equation 2x2 - kx+ k +2 = 0 is 3:2
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
No assumption of factorability for the given equation:
Roots using general solution formula,
x=%28k%2B-+sqrt%28k%5E2-4%2A2%28k%2B2%29%29%29%2F%282%2A2%29

%28k%2B-+sqrt%28k%5E2-8k-16%29%29%2F%282%2A2%29

cross%28%28k%2B-+sqrt%28%28k-4%29%5E2%29%29%2F%282%2A2%29%29
cross%28x=%28k%2B-+%28k-4%29%29%2F4%29


Using the given ratio for the roots 3:2 the resulting ratio equation is
-
%28k%2Bsqrt%28k%5E2-8k-16%29%29%2F%28k-sqrt%28k%5E2-8k-16%29%29=3%2F2
and algebraic steps lead to
3k%5E2-25k-50=0

Find discriminant, 25%5E2%2B4%2A3%2A50=1225=35%5E2

General solution formula to get k
k=%2825%2B-+35%29%2F6

highlight%28system%28k=-5%2F3%2COR%2Ck=10%29%29




----------Mistake early in the steps makes all of the below wrong----------------
Work through the quadratic equation and find that x=%28k-1%2B-+sqrt%28k%5E2-2k-15%29%29%2F4.
Setup the next equation according to the ratio of the roots.
%28k-1%2Bsqrt%28k%5E2-2k-15%29%29%2F%28k-1-sqrt%28k%5E2-2k-15%29%29=3%2F2
Simplify that and reach the equation 6k%5E2-12k-47=0.
If no mistakes were made in getting to this, then solve this quadratic equation for k. Discriminant is 1272.
General solution formula for quadratic equation results in
k=6%2B-+sqrt%28318%29.

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
The solution by "josgarithmetic" is  TOTALLY  and  ABSOLUTELY   W R O N G .

Below find the correct solution. 


Let "a" be one solution of the given equation  2x%5E2+-+kx%2B+k+%2B2 = 0.

Then the other solution is %283%2F2%29%2Aa.


If so, then the given polynomial 2x%5E2+-+kx%2B+k+%2B2 admits factorization

2x%5E2+-+kx%2B+k+%2B2 = 2%2A%28x-a%29%2A%28x-%283%2F2%29a%29 = (x-a)*(2x-3a) = 2x%5E2+-+5ax+%2B+3a%5E2.


It means that  -k = -5a, 3a^2 = k+2,   or  (substituting)

5a+%2B+2 = 3a%5E2,

3a%5E2+-+5a+%2B2 = 0  --->  a%5B1%2C2%5D = %285+%2B-+sqrt%2825+-+4%2A3%2A2%29%29%2F%282%2A3%29 = %285+%2B-+1%29%2F6.


1.  a%5B1%5D = 5%2B1%29%2F6 = 1  ---->  k = 5a = 5;


2.  a%5B2%5D = %285-1%29%2F6 = 4%2F6 = 2%2F3.  ---->  k = 5a = 10%2F3.


Answer.  There are TWO solutions for k:  1) k = 5,   and   2)  k = 10%2F3.

Solved.


God save you of using the method that "josgarithmetic" uses.