Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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. –  NikolajK 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
add comment

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 $$t<-2$$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.