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Find the set of values of k for which this equation has two real roots.

9x^2 + 2kx + k = 0

 Dec 1, 2020
 #1
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\(\frac{-k-\sqrt{k^2-9k}}{9}\)

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 Dec 1, 2020
 #2
avatar+114592 
+1

Using dududsu2's idea....

 

If we have real roots, the discriminant must  be  ≥  0

 

So

 

(2k)^2  - 4 (9)k   ≥  0

 

4k^2  - 36k  ≥  0

 

k^2 - 9k   ≥ 0

 

k ( k - 9) ≥  0

 

The solutions  for k  are    (-inf, 0]    and  [ 9, inf )

 

 

cool cool cool

 Dec 1, 2020

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