SOLUTION: For what values of k does 2x^2 - 5x - k = 0 have no real roots? Can you please help me out? Thanks so much in advance:)

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Question 825876: For what values of k does 2x^2 - 5x - k = 0 have no real roots?
Can you please help me out? Thanks so much in advance:)
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+2x%5E2+-+5x+-+k+=+0+
Use the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+2+
+b+=+-5+
+c+=+-k+
the roots are imaginary when +b%5E2+-+4%2Aa%2Ac+
is negative. Since +b%5E2+ is always positive.
+4%2Aa%2Ac+%3E+b%5E2+
+4%2A2%2A%28-k%29+%3E+%28-5%29%5E2+
+-8%2Ak+%3E+25+
+k+%3C+-25%2F8+ answer
( note that I had to reverse the inequality sign
when I divided by a negative number )
---------------
Suppose +k+=+-25%2F8+
+b%5E2+-+4%2Aa%2Ac+=+%28-5%29%5E2+-+4%2A2%2A%28-k%29+
+b%5E2+-+4%2Aa%2Ac+=+25+-+4%2A2%2A%28-%28-25%2F8%29%29+
+b%5E2+-+4%2Aa%2Ac+=+25+-+25+
+b%5E2+-+4%2Aa%2Ac+=+0+
That means there is 1 real root
---------------------------
Suppose +k+=+-26%2F8+ or
+k+=+-13%2F4+
+b%5E2+-+4%2Aa%2Ac+=+%28-5%29%5E2+-+4%2A2%2A%28-k%29+
+b%5E2+-+4%2Aa%2Ac+=+25+-+4%2A2%2A%28-%28-13%2F4%29%29+
+b%5E2+-+4%2Aa%2Ac+=+25+-+26+
+b%5E2+-+4%2Aa%2Ac+=+-1+
That means there are 2 imaginary roots

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

For what values of k does 2x^2 - 5x - k = 0 have no real roots?
Can you please help me out? Thanks so much in advance:) 


For values of k to have no real roots, the DISCRIMINANT MUST BE < 0 (negative). Therefore, we set the

DISCRIMINANT as negative, and solve. 

DISCIMINANT: b%5E2+-+4ac, with: a = 2   ;  b = - 5  ;  c = - k

Then, b%5E2+-+4ac+%3C+0 becomes:

      %28-+5%29%5E2+-+4%282%29%28-+k%29+%3C+0

      25+%2B+8k+%3C+0

      8k+%3C+-+25 

      highlight_green%28k+%3C+-+25%2F8%29

You can do the check!! 



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