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Find the set of possible values of k for each equation with the roots described.

1) x^2 - kx + k = 0 has two distinct real roots

2) 2x^2 + kx + k = 0 has no real roots 

3) x^2 + (k+4) x+(4K+1) = 0 has real roots 

4) (k-1) x^2 +(3k-1) x+ (k+5) = 0 has no real roots 

Guest Sep 20, 2018
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1) x^2 - kx + k = 0 has two distinct real roots

 

This will occur when the discriminant  is  > 0

So

 

(-k)^2  - 4 (1)(k)  > 0

k^2  - 4k > 0

k (k - 4) > 0

 

This will be true when  k < 0   or when k > 4

 

 2)  2x^2 + kx + k = 0 has no real roots

 

This will occur when the discriminant is < 0

So

 

k^2  - 4(2)(k)  < 0

k^2 - 8k < 0

k(k - 8) < 0

 

This will be true when  k lies in the interval  (0, 8)

 

3) x^2 + (k+4) x+(4K+1) = 0 has real roots 

 

This will occur when the discriminant = 0  (double real root)  or the discriminant > 0  [two distinct real roots ]

So

 

(k + 4)^2  -  4 (1) (4k + 1)  = 0

k^2 + 8k + 16  - 16k - 4  = 0

k^2 - 8k + 12  = 0

(k - 6) (k - 2)  = 0

Setting both factors to 0  and solving for  k  produces  k = 6  or k = 2

 

If the discriminant > 0...we arrive at

(k - 6) (k - 2) > 0

This will be true whenever k < 2  or when k > 6

 

 

 

4) (k-1) x^2 +(3k-1) x+ (k+5) = 0 has no real roots

 

This will be true when the discriminant < 0

So

 

(3k - 1)^2 - 4(k - 1)(k + 5)  < 0

9k^2 - 6k + 1 - 4 (k^2 + 4k - 5) < 0

9k^2 - 6k + 1 - 4k^2 - 16k + 20 < 0

5k^2 - 22k + 21 < 0

(5k - 7)(k - 3) < 0

 

This will be true whenever  k  lies in the interval  (7/5, 3)

 

 

 

cool cool cool

CPhill  Sep 20, 2018

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