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Really hard time with discriminants.

 Sep 26, 2017
 #1
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Here are the rules

(1) If the discriminant  = 0.....only one real  [double] root 

(2) If the discriminant > 0......two real roots

(3) If the discriminant < 0, no real roots

 

x^2 - kx + 4  = 0

 

If this has equal roots , the discriminant  = 0...so

 

b^2  - 4ac  = 0

 

(-k)^2  - 4(1)(4)  = 0

k^2  - 16  = 0 

k^2  = 16        taking both roots....k  = 4    or  k  =  -4

Prove this  to yourself     x^2 - 4x  + 4  = 0 →     (x - 2)^2  = 0       x  = 2 is a [double] root    or

x^2  + 4x + 4  =  0  →    (x + 2)^2  =0      and x  = -2   is a [double  root ]   

 

 

 

x^2  - 6x  + k  =0        equal roots....so....

 

(-6)^2  - 4 (1)k  = 0

36  - 4k  = 0    add 4k to both sides

36  = 4k         divide both sides by 4

k  = 9

Proof    x^2  - 6x  + 9   = 0    →  ( x - 3)^2   =  0    →    x  = 3  is a [double] root

 

 

 

3x^2 -2x + k  = 0 

So...the discriminant  must be  <  0

 

So

 

(-2)^2  - 4 (3)k  < 0

4  - 12k < 0     add 12k to both sides

4  < 12k          divide both sides by 12

(1/3) < k    →    k > 1/3

Prove this......look at the graph here :  https://www.desmos.com/calculator/cffskj6a9v

When  k  = 1/3....we have a [ double] root at x  = 1/3

But....when, for instance, k  =2/3, notice that we have no real roots

So any  k > 1/3    will produce imaginary  roots

 

 

 

cool cool cool

 Sep 26, 2017

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