Intermediate Algebra

[jamjam]

This kind of equation is new to me, did I get the correct answer?
9/x - 2 = 5/x^2
x =9 + sq rt 41, 9 - sq rt 41
x rationalized as 3.85,0.65
x rationalized as 15.4

 

[wjm11]

This kind of equation is new to me, did I get the correct answer?
9/x - 2 = 5/x^2
x =9 + sq rt 41, 9 - sq rt 41

You didn't show your work, but I'll assume you used the quadratic formula.

What happened to your "2a" denominator? Your answers should be over 4.

 

[mmm4444bot]

?

9/x - 2 = 5/x^2

x = 9 + sq rt 41, 9 - sq rt 41 Bill is correct. You forgot to type the denominators.

x rationalized as 3.85,0.65 "Rationalized" is not the correct verb; these numbers are "approximated".

x rationalized as 15.4 This number is not a solution.

The solutions are:

\(\displaystyle x \ = \ \frac{9}{4} \ + \ \frac{\sqrt{41}}{4} \ \approx \ 3.85\)

OR

\(\displaystyle x \ = \ \frac{9}{4} \ - \ \frac{\sqrt{41}}{4} \ \approx \ 0.65\)

?

 

[wannabeanerd]

mmm4444bot Is correct

 

[jamjam]

This kind of equation is new to me, did I get the correct answer?
9/x - 2 = 5/x^2
x =9 + sq rt 41, 9 - sq rt 41

You didn't show your work, but I'll assume you used the quadratic formula.

What happened to your "2a" denominator? Your answers should be over 4.

Thank you for bringing this to my attention and helping me with this problem.

 

[jamjam]

Re:

?

9/x - 2 = 5/x^2

x = 9 + sq rt 41, 9 - sq rt 41 Bill is correct. You forgot to type the denominators.

x rationalized as 3.85,0.65 "Rationalized" is not the correct verb; these numbers are "approximated".

x rationalized as 15.4 This number is not a solution.

The solutions are:

\(\displaystyle x \ = \ \frac{9}{4} \ + \ \frac{\sqrt{41}}{4} \ \approx \ 3.85\)

OR

\(\displaystyle x \ = \ \frac{9}{4} \ - \ \frac{\sqrt{41}}{4} \ \approx \ 0.65\)

?

Thank you for correcting me on this and helping me with this problem