AQA AS Maths question from '09 paper

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The discussion revolves around solving a quadratic equation to determine conditions for real roots. The equation (k+1)x² + 12x + (k-4) = 0 leads to the conclusion that k² - 3k - 40 ≤ 0 must hold true for real roots to exist. Participants clarify the values of coefficients a, b, and c, emphasizing that a = k+1 and c = k-4. The condition for real roots is explained using the discriminant formula b² - 4ac, which must be greater than or equal to zero. The conversation concludes with a participant expressing gratitude for the clarification received.
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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 alot, I really appreciate it! :smile:
 
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