# Homework Help: Nature of roots of quadratic equations

1. Dec 10, 2011

### thornluke

1. The problem statement, all variables and given/known data
The equation kx2 - 3x + (k+2) = 0 has two distinct real roots. Find the set of possible values of k.

2. Relevant equations
Since the equation has two distinct real roots, b2 - 4ac > 0

3. The attempt at a solution
b2-4ac>0
9-4(k+2)(k)>0
9-4(k2+2k) >0
9-4k2-8k>0
= -4k2-8k+9>0
Multiply both sides by -1,
4k2+8k-9>0
(4k-3)(k+3)>0

-3<k<3/4

2. Dec 10, 2011

### SammyS

Staff Emeritus

(4k-3)(k+3) = 4k2 + 9k - 9 .

Also, multiplying by -1 will change > to < .

Solve 4k2+8k-9 = 0 by using the quadratic formula --- or by completing the square.

3. Dec 10, 2011

### thornluke

((-8 ± √208)/8) < 0

-2.80 < k < 0.80

I'm getting closer to the "answer" (-2.46<k<0.458) am I wrong, or is the "answer" wrong?

4. Dec 10, 2011

### eumyang

I'm getting the same roots (-2.80 and 0.80).

When I played around with the coefficients of the original quadratic, I found that if you made the coefficient of the x2 term 2k:
2kx2 - 3x + (k+2) = 0
You will get the original answer that you stated: -2.4577 < k < 0.4577. So it looks like either you copied the problem incorrectly or the book has a typo somewhere.

5. Dec 10, 2011

### SammyS

Staff Emeritus
The text book's answer is consistent with -8k2 - 16k +9 > 0 . equivalent to -4k2 - 8k + 9/2 >0

It's hard to see how that's from a simple Typo -- unless the coefficient of x is should have been 3/√2 in the initial equation.

6. Dec 12, 2011

### thornluke

I guess everyone makes mistakes.. it gets extremely annoying when textbooks provide you with the wrong answers or have a typo.

7. Dec 13, 2011