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up vote 3 down vote favorite

It says:

Solve the parameter $t$ when the equation $x^2=2t+4$ hasn't any real roots

And well... I could solve for $t$, for $x$... but I don't get the actual point of the question. What should I do there?

Thank you

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I could imagine that the task is to find the set of t-values, for which there are no real x-values (or the other way around). But the english is strange. – Nick Kidman Dec 30 '11 at 3:27
Suppose $t=-10$, can you find $x$ such that $x^2 = 2t+4$? Why? – sdcvvc Dec 30 '11 at 3:33

1 Answer

up vote 4 down vote accepted

If the equation has no real roots, then what it means is that if you solve for $x$ you have to get complex values for $x$. In order for this to happen what is on the right hand side must be negative since the square root of negative numbers is not real but complex.

This means you have the following ;$$2t+4<0$$ which implies that $$t<-2$$

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I don't get that... – Anna Dec 30 '11 at 3:29
1  
Anna, what do you get? Do you know what the word "real" means in this problem? – Gerry Myerson Dec 30 '11 at 3:31
Ahhh, I just got it. Thank you :) – Anna Dec 30 '11 at 3:35

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