Hi Everyone,
need some help to transpose (attached picts)Thanks very much in advance.
Hi vb14, :)
Welcome to MHB! The first equation can be simplified to,
$$q=\sqrt{6p^2-4}$$
Now square both sides to get,
$$q^2=6p^2-4$$
Add $4$ to both sides,
$$q^2+4=6p^2$$
Divide both sides by $6$,
$$\frac{q^2+4}{6}=p^2$$
Take the square root of both sides,
$$p=\pm\sqrt{\frac{q^2+4}{6}}$$
Can you give it a try for the second one?
#3
vb14
8
0
can you give a hand with second one as well?
please:)
#4
MarkFL
Gold Member
MHB
13,288
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Start by cubing both sides to get rid of the rational exponent on the right side...what do you have now?
#5
vb14
8
0
2x-5x2=(6x2+5)/t3
#6
MarkFL
Gold Member
MHB
13,288
12
Where did the $x$ on the left side come from?
I would first just write:
\(\displaystyle t^3=\frac{6x^2+5}{2-5x^2}\)
Next, multiply through by $2-5x^2$ to get:
\(\displaystyle 2t^3-5t^3x^2=6x^2+5\)
Now, what you want to do is arrange the equation with all the terms involving $x$ on one side and everything else on the other, the factor out $x^2$ and divide through by the other factor to solve for $x^2$ and then take the square root of both sides.
edit: I see you have also posted this at MMF and have been given the solution.
#7
vb14
8
0
got this, is it correct?
Attachments
Transpose.PNG
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#8
MarkFL
Gold Member
MHB
13,288
12
Yes, that is correct. Do you understand how that was obtained?