Limits question + polynomials question (1 Viewer)

Zeber

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2x^2+6x+1+kx^2+2k
= (2+k)x^2 + 6x + 1+2k
do the discriminant < 0 and k > -2 as 2+k > 0 for all x to be positive.
 
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Iruka

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for the limit q, multiply both numerator and denominator by sqrt(20-x) +5, then simplify.
 

gurmies

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2x^2 + 6x + 1 + k(x^2+2)= 0
2x^2 + 6x + kx^2 + 2k = 0
(k+2)x^2 + 6x + 1 + 2k = 0
Now two conditions, discriminant has to be < 0 and coefficient has to be positive. So for now we can say that k > -2.

Now examining the disciminant b^2 - 4ac.

36 - 4(k+2)(2k+1) < 0

Solving for 36 - 4(k+2)(2k+1) = 0 (known as the gate post method, it's my preferred way of doing inequality crap, but you can do it however you like)

9 - (k+2)(2k+1) = 0 (divide both sides by 4)

9 - (2k^2 + k + 4k + 2) = 0

9 - 2k^2 - k - 4k - 2 = 0

2k^2 + 5k - 7 = 0

(2k+7)(k-1) = 0

k = -7/2 or k = 1

Since we already said k > -2 to satisfy the coefficient condition, we can drop the answer -7/2. Now we test the two domains, -2 < k < 1 and k > 1. Turns out that k must be greater than 1 to satisfy the discriminant and thus, as it also satisfies the coefficiant condition, that is the restriction. k > 1. Zeber I think you're not quite correct on that one. Try k = 1/2, and the discriminant is > 0
 
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Zeber

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If you read my post properly, I told him to do the discriminant < 0 and consider k > -2 as the k+2 > 0.
 

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