Evaluate the expression and write the result in the form .
The real number a equals ?
The real number b equals ?
First look for the critical points. This is where the expression is 0 or where a 0 appears in the denominator. Since we have no denominator, all we need to know is where the expression is 0. So:
So the critical points are at x = -5 and x = 6.
Now split the real line into intervals and test each interval.
: Yes!
: No.
: Yes!
So the solution set is .
-Dan
2 last things. Is this right?
If you rationalize the denominator of
then you will get
where r, s, and n are all positive integers (with no common factors).
r= -7
s= 4
n=-118
And the last question I have no clue how to do.
Find all solutions of the equation and express them in the form :
First input the solution with b<0 here:
the real number a equals _____ and the real number b equals ______
Then input the solution with b>0 here:
the real number a equals _______ and the real number b equals _________
[quote=badandy328;28061]2 last things. Is this right?
If you rationalize the denominator of
then you will get
where r, s, and n are all positive integers (with no common factors).
r= -7
s= 4
n=-118
[\quote]
Almost. The conjugate of is . You only negate one of the terms. Think of it this way. We are applying the identity: . This ensures that the square root signs disappear.
Multiply both sides by t:
<-- We have to remember to check the solutions and make sure because the original expression forbids it. As it happens we won't have to worry here.
Using the quadratic formula:
-Dan
This one:
And the last question I have no clue how to do.
Find all solutions of the equation and express them in the form :
First input the solution with b<0 here:
the real number a equals _____ and the real number b equals ______
Then input the solution with b>0 here:
the real number a equals _______ and the real number b equals _________[/QUOTE]