Question 481782: hi i need some help. here is the qns
Find the values of k for which the quadratic equation 9x^2-kx+(k-7)=0 has
(i) one positive and one negative root
(ii) one root twice the value of the other
Answer by mathstutor494(120)   (Show Source):
You can put this solution on YOUR website! 9x^2-kx+(k-7)=0
1. Condition for a quadratic equation ax^2+bx+c=0 to have one positive and one negative root is that discriminant b^2-4ac > 0
Substituting values of a, b & c from the problem, we get
(-k)^2 - 4*9*(k-7)>0
k^2- 36k + 252 >0
Solving above quadratic equation results in k= 42, -6
Hence k=42
2. Sum of the roots S = x1+x2 = -b/a and product of roots P=x1*x2=c/a
Here x1 = x and x2 = 2x
So S = x+2x = -(-k)/9 and P = x*2x = (k-7)/9
Accordingly S= 3x = k/9 .........(1)
and
P = 2x^2 = (k-7)/9................(2)
Substituting value of x from eqn 1 in 2
2*(k/27)^2 = (k-7)/9
18k^2 = 729(k-7)
Transposing...
18k^2 - 729k + 5103 = 0
Solving for k results in
k = 315, 9
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