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Need help to solve this question

I was wondering how to solve this following equation:
0.075=(B^2)*sqrt(1.2-B/2)

Thanks very much.

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Reply 1
Avatar for Dr A
Dr A
Original post by cwynter
I was wondering how to solve this following equation:
0.075=(B^2)*sqrt(1.2-B/2)

Thanks very much.


You have

0.075=b21.2−b20.075 = b^2 \sqrt{1.2-\frac{b}{2}}

How about squaring both sides? (square roots are generally awkward).

Also, 0.075=3400.075 = \frac{3}{40}
(edited 11 years ago)
Reply 2
Avatar for Madalaine M
Madalaine M
Rubbish. Deleted.
(edited 11 years ago)
Reply 3
Avatar for cwynter
cwynter
OP
I tried that, but got confused for solving B^4 and B^5
9/800=(2.4*B^4)-(B^5)

Madalaine M-That's very clear. Thanks a lot.
(edited 11 years ago)
Reply 4
Avatar for Dr A
Dr A
Original post by Madalaine M
This is my method:

0.075=b21.2−b20.075=b^2 \sqrt{1.2-\frac{b}{2}}

divide both sides by b2b^2

0.075b2=1.2−b2\frac{0.075}{b^2}=\sqrt{1.2-\frac{b}{2}}

square root both sides:

0.075b=1.2−b2\frac{\sqrt{0.075}}{b}=1.2-\frac{b}{2}

Times both sides by 2:

20.075b=2.4−b\frac{2\sqrt{0.075}}{b}=2.4-b

Times both sides by bb

20.075=2.4b−b22\sqrt{0.075}=2.4b-b^2

Solve the resulting quadratic equation:

b2−2.4b+20.075=0b^2-2.4b+2\sqrt{0.075}=0

Can you solve this by yourself?


BAD ERROR - line 3 of working.



a=b  ⟹  a2=ba=\sqrt{b} \implies a^2 = b
(edited 11 years ago)
Reply 5
Avatar for Dr A
Dr A
Original post by cwynter
I tried that, but got confused for solving B^4 and B^5
9/800=(2.4*B^4)-(B^5)

Madalaine M-That's very clear. Thanks a lot.


There's a bad error in her working.
Reply 6
Avatar for Madalaine M
Madalaine M
Original post by cwynter
I tried that, but got confused for solving B^4 and B^5
9/800=(2.4*B^4)-(B^5)

Madalaine M-That's very clear. Thanks a lot.


Is it correct? You can just square root both sides without dividing both sides by b at the start.
Reply 7
Avatar for cwynter
cwynter
OP
Original post by Dr A
There's a bad error in her working.


Before line 3, her method is similar to what you suggested, which brings me back to the same problem: how to solve 9/800=(2.4*B^4)-(B^5)?
Reply 8
Avatar for Madalaine M
Madalaine M
Original post by TenOfThem
of course it is not correct

check again


Sorry, I meant square both sides to eliminate the square root.
Reply 9
Avatar for Madalaine M
Madalaine M
Original post by TenOfThem
sure

but you did not do that either

you rooted the LHS and squared the RHS


I know! It was rubbish. I'm tired and forgot to check it through.
Reply 10
Avatar for Dr A
Dr A
Original post by cwynter
Before line 3, her method is similar to what you suggested, which brings me back to the same problem: how to solve 9/800=(2.4*B^4)-(B^5)?


181600b4=2.4−b\displaystyle \frac{18}{1600b^4} = 2.4 - b

Go from there.
Reply 11
Avatar for cwynter
cwynter
OP
Original post by Madalaine M
I know! It was rubbish. I'm tired and forgot to check it through.

I'm just glad that at least you guys tried to help. Thanks for the effort anyway :smile:
Original post by Madalaine M
I know! It was rubbish. I'm tired and forgot to check it through.


Easily done

Especially when you have been revising a lot
Reply 13
Avatar for cwynter
cwynter
OP
Original post by Dr A
181600b4=2.4−b\displaystyle \frac{18}{1600b^4} = 2.4 - b

Go from there.


I'm sorry but I still don't see it.
Original post by TenOfThem
Easily done

Especially when you have been revising a lot


That reminds me, can you help me with a C3 question I'm struggling with?
Original post by Madalaine M
That reminds me, can you help me with a C3 question I'm struggling with?


what
Reply 16
Avatar for Dr A
Dr A
18=3840b4−1600b518=3840b^4 -1600b^5

1600b5−3840b4+18=01600b^5 - 3840b^4 + 18=0
(edited 11 years ago)
Original post by Dr A
18=3840b4−1600b518=3840b^4 -1600b^5

1600b5−3840b4+18=01600b^5 - 3840b^4 + 18=0

Use

u=b4u = b^4

to obtain

1600u2−3840u+18=01600u^2 - 3840u + 18=0

which has real roots.


This is not correct either as (b4)2≠b5(b^4)^2 \not = b^5
Reply 18
Avatar for Dr A
Dr A
Original post by TenOfThem
This is not correct either as (b4)2≠b5(b^4)^2 \not = b^5


Grrr... I think I better catch up on some sleep.
Original post by TenOfThem
what


Never mind. I figured it out as I was typing it out to you.

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