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Nature of roots of quadratic equations 
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#1
Dec1011, 01:18 AM

P: 37

1. The problem statement, all variables and given/known data
The equation kx^{2}  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, b^{2}  4ac > 0 3. The attempt at a solution b^{2}4ac>0 94(k+2)(k)>0 94(k^{2}+2k) >0 94k^{2}8k>0 = 4k^{2}8k+9>0 Multiply both sides by 1, 4k^{2}+8k9>0 (4k3)(k+3)>0 3<k<3/4 However the answer is 2.46<k<0.458 I'm lost, help please! 


#2
Dec1011, 01:48 AM

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(4k3)(k+3) = 4k^{2} + 9k  9 . Also, multiplying by 1 will change > to < . Solve 4k^{2}+8k9 = 0 by using the quadratic formula  or by completing the square. 


#3
Dec1011, 02:03 AM

P: 37

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
Dec1011, 06:11 AM

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Nature of roots of quadratic equations
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 x^{2} term 2k: 2kx^{2}  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
Dec1011, 02:32 PM

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The text book's answer is consistent with 8k^{2}  16k +9 > 0 . equivalent to 4k^{2}  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
Dec1211, 04:36 AM

P: 37




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