Nature of roots of quadratic equations

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Homework Help Overview

The discussion revolves around the quadratic equation kx² - 3x + (k+2) = 0, specifically focusing on determining the values of k for which the equation has two distinct real roots. Participants are analyzing the conditions under which the discriminant is positive.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply the discriminant condition (b² - 4ac > 0) to derive inequalities for k. There are discussions about the correctness of factoring and the implications of multiplying by negative values on inequalities. Some participants are also exploring variations of the original equation to see how changes affect the roots.

Discussion Status

There is an ongoing examination of the calculations and interpretations of the discriminant. Some participants are questioning the accuracy of the textbook's provided answer and suggesting that there may be a typo in the problem statement. Multiple interpretations of the problem are being explored, but no consensus has been reached.

Contextual Notes

Participants note potential discrepancies in the problem statement and the textbook answer, leading to confusion about the correct values of k. There is an acknowledgment of the possibility of errors in the original problem setup or in the textbook itself.

thornluke
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Homework Statement


The equation kx2 - 3x + (k+2) = 0 has two distinct real roots. Find the set of possible values of k.


Homework Equations


Since the equation has two distinct real roots, b2 - 4ac > 0

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

However the answer is -2.46<k<0.458
I'm lost, help please! :confused:
 
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thornluke said:

Homework Statement


The equation kx2 - 3x + (k+2) = 0 has two distinct real roots. Find the set of possible values of k.


Homework Equations


Since the equation has two distinct real roots, b2 - 4ac > 0

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

However the answer is -2.46<k<0.458
I'm lost, help please! :confused:
Your factoring is incorrect.

(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.
 
SammyS said:
Your factoring is incorrect.

(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.

((-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?
 
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.
 
The textbook'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.
 
eumyang said:
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.

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

They are put there to help you to build self-confidence. :biggrin:
 

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