Math Is Fun Forum

mathnovice1

Member

Registered: 2008-09-30

Posts: 1

2° degree equations/inequalities with square root

I'm sorry to bother.

Could you please fully explain to me what is the right approach to solving these equations/inequalities? (I didn't understand it in class today):

sqrt(x^2-1)=-1

and

sqrt(x^2-1)>=x

thank you so much

Chewy

Member

Registered: 2008-08-07

Posts: 67

Re: 2° degree equations/inequalities with square root

That doesn't make sense to me, because any positive value for x will result in a number greater than -1. Any negative value for x will result in the same. Any fractional value for x will result in a number other than -1. Maybe I am way off, but I just don't see what the solution is, other than :

sqrt(x^2-1) = sqrt [(x+1) * (x-1)]

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: 2° degree equations/inequalities with square root

In both cases you want to square both sides.  This is legal in the first case since even though you are multiplying by different expressions (sqrt(x^2-1) and -1) they are stated to be equal, so you're really performing the same operation to both sides.  You do want to be careful in the second case since negative values can switch the direction of the less than sign.  You should state separate cases for when the expressions are greater than 0 and when they're less than 0.

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