SOLUTION: What is the largest value of k such that the equation x2 – 5x + k has at least one real root?

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Question 922555: What is the largest value of k such that the equation x2 – 5x + k has at least one real root?
Found 2 solutions by jim_thompson5910, josmiceli:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-5x%2Bk is in the form ax%5E2%2Bbx%2Bc where a+=+1, b=-5,c+=+k


D+=+b%5E2+-+4ac


D+=+%28-5%29%5E2+-+4%2A1%2Ak Plug in a+=+1, b=-5,c+=+k


D+=+25+-+4k


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We will have at least one real solution when D+%3E=+0


D+%3E=+0


25+-+4k+%3E=+0 Plug in D+=+25+-+4k. Solve for k.


25+%3E=+4k


4k+%3C=+25


k+%3C=+25%2F4


k+%3C=+6.25


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The answer as a fraction is k+%3C=+25%2F4


The answer as a decimal is k+%3C=+6.25


If you insist on having k be a whole number, then round to the nearest whole number to get k+=+6


In terms of a decimal, the largest k can be is 6.25 or 6 (depending on if you want a whole number or not).



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Let me know if that helps or not. Thanks.

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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E2+-+5x+%2Bk+
The quadratic formula is:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
The "discriminant" is:
+b%5E2+-+4a%2Ac+
For at least 1 real root,
+b%5E2+%3E=+4%2Aa%2Ac+
+b+=+-5+
+a+=+1+
+c+=+k+
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+%28-5%29%5E2+%3E=+4%2A1%2Ak+
+25+%3E=+4k+
+k+%3C=+25%2F4+
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check:
suppose +k+=+25%2F4+
+b%5E2+-+4a%2Ac+=+%28-5%29%5E2+-+4%2A1%2A%2825%2F4%29+
+b%5E2+-+4a%2Ac+=+25+-+25+
+b%5E2+-+4a%2Ac+=+0+
which means 1 real root
and if +k%3E+25%2F4+, there are 2 real roots