AQA AS Maths question from '09 paper

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Discussion Overview

The discussion revolves around a quadratic equation from an AS Maths exam paper, specifically focusing on the conditions for the equation to have real roots. Participants explore how to derive the inequality k² - 3k - 40 ≤ 0 from the equation (k+1)x² + 12x + (k-4) = 0.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to approach the problem and seeks assistance.
  • Another participant suggests using the discriminant condition b² - 4ac ≤ 0 to determine the nature of the roots, substituting the values of a, b, and c from the quadratic equation.
  • There is a moment of uncertainty regarding the constant term in the inequality, with a participant initially questioning whether it should be -40 or -140.
  • Clarification is provided on how to identify the coefficients a and c, with a being k+1 and c being k-4.
  • Participants discuss the implications of the discriminant being greater than or equal to zero for the existence of real roots, emphasizing the need for the derived inequality.

Areas of Agreement / Disagreement

Participants generally agree on the method of using the discriminant to analyze the roots of the quadratic equation. However, there is some initial confusion regarding the constants and the formulation of the inequality, which is clarified through the discussion.

Contextual Notes

Some participants express uncertainty about the steps involved in deriving the inequality and the definitions of the coefficients, indicating a need for further clarification on these mathematical concepts.

Physssics
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Hi,
I have my AS Maths exam on Monday and so I was looking at some previous papers and I couldn't do this question:

The quadratic equation (k+1)x² + 12x + (k-4) = 0 has real roots
(a) Show that k² - 3k - 40 ≤ 0

I don't even understand where to start so I was hoping you could help me.

Thanks!
 
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I think it may be, that as it has real roots, it must mean that b^2-4ac ≤ 0
so just substitute your values of a b and c into the formula to get k² - 3k - 40

are you sure its -40 and not -140 ?
cause then it would work i think!
 
oh no sorry, i forgot to multiply it out, silly me, it would work with 40.
and also 0 ≤ b^2-4ac if it has real roots. sorry I'm a bit foggy on this stuff!
 
Can I ask how how you got the values for a and c?
a=
b= 12
c=
Because I don't understand why the x is there?

By the way -- thanks for the response, I really really appreciate it!
 
well you don't have any integer values for the a and c
so instead, you just substitue in the constants before the x^2 for a, and the last value for c

so a would simply be k+1
and c would be k-4

do you remember how to find if an equation has real roots ? that is where the b^2-4ac comes in, you could google it if you don't. If the answer is less than zero it has no real roots, equal to zero it has one real root etc. But yours is telling you it has real roots. plural. so therefore your answer to b^2-4ac is, it must be bigger than 0. Your answer will include k once you multiply it out, and you should get the equation you stated before
k² - 3k - 40 ≤ 0
 
Dollydaggerxo said:
well you don't have any integer values for the a and c
so instead, you just substitue in the constants before the x^2 for a, and the last value for c

so a would simply be k+1
and c would be k-4

do you remember how to find if an equation has real roots ? that is where the b^2-4ac comes in, you could google it if you don't. If the answer is less than zero it has no real roots, equal to zero it has one real root etc. But yours is telling you it has real roots. plural. so therefore your answer to b^2-4ac is, it must be bigger than 0. Your answer will include k once you multiply it out, and you should get the equation you stated before
k² - 3k - 40 ≤ 0

Yeah, I understand now. Thanks a lot, I really appreciate it! :smile:
 

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