# Algebra

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#1
f(x) = x^2 + (k+3)x +k

How do I show for all values of K the equation of f(x) = 0 hs real roots.

I have found the discriminant and put it in this format:

k^2 +2k +9
(k+1)^2 +8 = 0

but I don't remember what I've got to do with this now to answer the question. Please, someone, help it would really be appreciated
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1 month ago
#2
(Original post by Hollymae764)
f(x) = x^2 + (k+3)x +k

How do I show for all values of K the equation of f(x) = 0 hs real roots.

I have found the discriminant and put it in this format:

k^2 +2k +9
(k+1)^2 +8 = 0

but I don't remember what I've got to do with this now to answer the question. Please, someone, help it would really be appreciated
What does the discriminant have to be for f(x) to have real roots? How would you prove that k^2+2k+9 is this? You've pretty much already done it.
Last edited by Bill V2; 1 month ago
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#3
b^2-4ac> 0
so (k+1)^2 +8 >0
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#4
(Original post by Hollymae764)
b^2-4ac> 0
so (k+1)^2 +8 >0
What do I do next?
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1 month ago
#5
(Original post by Hollymae764)
What do I do next?
Whenever you square a (real) quantity, its always greater than or equal to zero, so can you reason from there?
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1 month ago
#6
(Original post by Hollymae764)
What do I do next?
Show that the expression is always positive, i.e. when you square a real number, it is positive, and so (k+1)^2 is positive
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