I'm stuck on this problem (please help) :(

[hannibolio]

Hello to all who read this thread. My friend asked me for help on some problems but I've reached a road bump. The problems ask "solve the equation" and are written as follows:

1) m - m^(1/2) - 12 = 0 ------> Answer: 16

2) x^4 - 11x^2 + 18 = 0 ------ Answer: plus and minus 2^(1/2) , plus and minus 3

3) x^1/2 - 2x^1/4 - 3 = 0 -----> Answer: 81

4) 10/p^2 + 3/p - 1 = 0 ------> Answer: 2 , 5

I have work for number 1, and 2 but number 3 and 4 I'm not sure how to do them. For number 1 and 2, I've tried the factor x (two numbers that multiply to equal the top number but add up to equal the bottom number), quadratic formula and I get nothing close to the answer Thanks to whoever can help me. I know I'm asking for a lot :???: so I really would appreciate the help. Thanks guys.

 

[JeffM]

Hello to all who read this thread. My friend asked me for help on some problems but I've reached a road bump. The problems ask "solve the equation" and are written as follows:

1) m - m^(1/2) - 12 = 0 ------> Answer: 16

2) x^4 - 11x^2 + 18 = 0 ------ Answer: plus and minus 2^(1/2) , plus and minus 3

3) x^1/2 - 2x^1/4 - 3 = 0 -----> Answer: 81

4) 10/p^2 + 3/p - 1 = 0 ------> Answer: 2 , 5

I have work for number 1, and 2 but number 3 and 4 I'm not sure how to do them. For number 1 and 2, I've tried the factor x (two numbers that multiply to equal the top number but add up to equal the bottom number), quadratic formula and I get nothing close to the answer Thanks to whoever can help me. I know I'm asking for a lot :???: so I really would appreciate the help. Thanks guys.

I shall show you to do the first one. Then you do the rest because they all involve the same idea. If you get stuck, come back and show your work.

\(\displaystyle Given\ m - m^{(1/2)} - 12 = 0.\)

\(\displaystyle Let\ y = m^{(1/2)} =\sqrt{m} \implies y^2 = m \implies\)

\(\displaystyle y^2 - y - 12 = 0 \implies (y - 4)(y + 3) = 0 \implies y = 4\ or\ y = - 3 \implies \sqrt{m} = 4\ or\ \sqrt{m} = - 3.\)

\(\displaystyle But\ there\ is\ no\ real\ number\ m\ such\ that\ \sqrt{m} = - 3 \implies\)

\(\displaystyle \sqrt{m} = 4 \implies m = 16.\)

Let's check.

\(\displaystyle 16 - \sqrt{16} - 12 = 16 - 4 - 12 = 12 - 12 = 0.\)

 

[hannibolio]

I shall show you to do the first one. Then you do the rest because they all involve the same idea. If you get stuck, come back and show your work.

\(\displaystyle Given\ m - m^{(1/2)} - 12 = 0.\)

\(\displaystyle Let\ y = m^{(1/2)} =\sqrt{m} \implies y^2 = m \implies\)

\(\displaystyle y^2 - y - 12 = 0 \implies (y - 4)(y + 3) = 0 \implies y = 4\ or\ y = - 3 \implies \sqrt{m} = 4\ or\ \sqrt{m} = - 3.\)

\(\displaystyle But\ there\ is\ no\ real\ number\ m\ such\ that\ \sqrt{m} = - 3 \implies\)

\(\displaystyle \sqrt{m} = 4 \implies m = 16.\)

Let's check.

\(\displaystyle 16 - \sqrt{16} - 12 = 16 - 4 - 12 = 12 - 12 = 0.\)

Thank you so much. You saved my life! I appreciate the help Jeff!

 

[hannibolio]

What happens if you multiply that equation by p^2? Here:
10 + 3p - p^2 = 0 ; get that?

Btw, the 2 you gave as one of answers should be -2.

I didn't even think to do that... Okay, so I did that and I got what you got,worked it out, and got an answer of -2. I found that p =5 also is an answer but I could be wrong. Here's my work:

10/p^2 + 3/p - 1 = 0

p^2 (10/p^2 + 3/p - 1) =p^2(0)

10 + 3p - p^2 = 0

-1(-p^2 +3p + 10) =-1(0)

p^2 - 3p - 10 = 0

(p + 2)(p - 5) = 0

P + 2 = 0
p = -2
----------
p - 5 = 0
p = 5
-------------------------------
check for p = -2
(-2+2)(-2 - 5) = 0
(0)(-7) = 0
0=0
-------------------------------
check for p = 5
(5+2)(5-5)=0
(7)(0)=0
0=0

 

[hannibolio]

Yep! Easy, right

Thanks again sir! I was wondering if you can give me a hint for #3? I'm doing it right now but I haven't made any progress with it. I've tried the quadratic formula, I don't think it can be factored. I've tried this though:

x^1/2 - 2x^1/4 - 3 = 0

( x^1/2 - 2x^1/4 - 3)^4 = (0)^4

x^2+16x+81=0
-------------------------
First method I've tried:

(x^2+4x+4)= -81+4

(x+2)^2 = -77---------> I chose to stop here because I got the imaginary number "i" on the right side.
--------------------------------------
So far, this is where I'm stuck.

 

[hannibolio]

Thanks again sir! I was wondering if you can give me a hint for #3? I'm doing it right now but I haven't made any progress with it. I've tried the quadratic formula, I don't think it can be factored. I've tried this though:

x^1/2 - 2x^1/4 - 3 = 0

( x^1/2 - 2x^1/4 - 3)^4 = (0)^4

x^2+16x+81=0
-------------------------
First method I've tried:

(x^2+4x+4)= -81+4

(x+2)^2 = -77---------> I chose to stop here because I got the imaginary number "i" on the right side.
--------------------------------------
So far, this is where I'm stuck.

Okay, solved it. I did u substitution and got x = plus or minus root 2 and plus or minus 3.

 

[hannibolio]

Okay, solved it. I did u substitution and got x = plus or minus root 2 and plus or minus 3.

Okay, solved all of them. U substitution was the way to go! Thanks to Jeff and Denis for the help. You guys are awesome! :-D Until we meet again sirs.