Possible Values of k for a Quadratic Inequality: 2x^2 + kx + 9 = 0

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SUMMARY

The discussion centers on determining the possible values of k in the quadratic equation 2x² + kx + 9 = 0, where one root is twice the other. Two methods were presented: one using the discriminant, yielding k > √72 or k < -√72, and another using the sum and product of roots, leading to k = ±9. The latter method is confirmed to be more complete as it directly verifies the relationship between the roots.

PREREQUISITES
  • Understanding of quadratic equations and their roots
  • Familiarity with the discriminant and its implications
  • Knowledge of the sum and product of roots formulas
  • Basic algebraic manipulation skills
NEXT STEPS
  • Explore the derivation of the quadratic formula and its applications
  • Learn about the discriminant and its role in determining the nature of roots
  • Study the relationships between roots and coefficients in polynomial equations
  • Investigate alternative methods for solving quadratic equations, such as factoring and completing the square
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Students studying algebra, particularly those focusing on quadratic equations, as well as educators looking for methods to teach root relationships and discriminant analysis.

LiHJ
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Homework Statement


Dear Mentors and Helpers,

Here's the question:
Find the possible values of k such that one root of the equation 2x^2 + kx + 9 = 0 is twice the other.

Homework Equations


My classmate's working:

Discriminate > 0
k^2 - (4)(2)(9) > 0
k^2 -72 > 0
[k + sqrt (72)] [k- sqrt(72)] > 0

Answer: k > sqrt (72) or k < - sqrt(72)

The Attempt at a Solution


My working:

Let p and 2p be the roots of the equation.

Sum of roots:
3p = (-k)/2
p = (-k)/6 -----(1)

Product of roots:
2p^2 = 9/2 ------(2)

Substitute (1) into (2):

2(-k/6)^2 = 9/2
(k^2)/36 = 9/4
k^2 = 81
k = +/- (9)
Answer: +/- (9)

Dear Mentors and Helpers,
Please help me to check whether my friends working is correct or mine is correct. Thanks for your time.

 
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LiHJ said:

Homework Statement


Dear Mentors and Helpers,

Here's the question:
Find the possible values of k such that one root of the equation 2x^2 + kx + 9 = 0 is twice the other.

Homework Equations


My classmate's working:

Discriminate > 0
k^2 - (4)(2)(9) > 0
k^2 -72 > 0
[k + sqrt (72)] [k- sqrt(72)] > 0

Answer: k > sqrt (72) or k < - sqrt(72)

The Attempt at a Solution


My working:

Let p and 2p be the roots of the equation.

Sum of roots:
3p = (-k)/2
p = (-k)/6 -----(1)

Product of roots:
2p^2 = 9/2 ------(2)

Substitute (1) into (2):

2(-k/6)^2 = 9/2
(k^2)/36 = 9/4
k^2 = 81
k = +/- (9)
Answer: +/- (9)

Dear Mentors and Helpers,
Please help me to check whether my friends working is correct or mine is correct. Thanks for your time.
I get the same thing you did. Your classmate's work is incomplete. All he did was to find intervals of k values for which the discriminant is positive.

You can take this problem one step further to verify that your work is correct, by finding the two roots of the quadratic. One of the roots should be twice the other. Note that there are two pairs of values that work.

Also, I approached this problem in a different way, since I don't have the sum of roots, product of roots formulas memorized.

I rewrote the equation as x2 + (k/2)x + 9/2 = 0. Then, since p and 2p are roots, it must be that (x - p)(x - 2p) = 0. Expand the second equation and then equate the coefficient of x and the constant term in the two equations. This will give you two equations in the unknowns k and p.
 
Thanks for the verification and further explanation of another method
 

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