how do u solve for r2 (squared)?

Question

1/R=1/r1+1/r2 solve for r2

r1=r to the first power.. and the little r and R and different.

Answer 1

Start by getting the same denominator for your fractions. To get the r to be r², you have to multiply top and bottom of 1/r by r. That makes it r/r²:

1 .... r .... 1
-- = --- + ---
R ... r² ... r²

Now add the fractions:
1 ....(r + 1)
-- = ---------
R ...... r²

Cross multiply:
r² = R(r + 1)

P.S. Are you *sure* those are exponents? They look like they might be subscripts. In other words r "one" and r "two", not r to the first power, r to the second power. Are they below the line?

If so, then you would use a similar method, but you can't use r² as the common denominator:

1 ... 1 .... 1
-- = --- + ---
R .. r1 ... r2

The common denominator would be r1 * r2, so multiply the first fraction on the right by r2 on top and bottom. And multiply the last fraction by r1 on top and bottom:
1 .... r2 .... r1
-- = ----- + ------
R .. r1r2 . r1r2

Now combine:
1 ... r1 + r2
-- = ----------
R .... r1 r2

Cross multiply:
r1 r2 = R(r1 + r2)

Distribute R through the parentheses on the right:
r1 r2 = R r1 + R r2

Subtract R r2 from both sides. This gets r2 all on one side:
r1 r2 - R r2 = R r1

Factor out the common r2:
(r1 - R) r2 = R r1

Divide both sides by (r1 - R)
........ R r1
r2 = ---------
.......(r1 - R)

Answer 2

I don't think the 1 in r1 is to be read as first power. I think it's a subscript. So I will interpret this way.

1/R=1/r1+1/r2 solve for r2
1/r2 = 1/R - 1/r1
Taking the reciprocal of both sides:
r2 = 1/[1/R - 1/r1] ANS

teddy boy

Answer 3

Your equation is equivalent to the quadratic equation
fro r:

r^2 - R r - R = 0 (eq 1)

The discriminant D of this equation is D=R^2 + 4 R, hence
we have 2 cases (I exclude of course the case R=0.)

Case 1) if D>0 which happens only if R >0 or R< - 4, the roots of eq 1are:

r+ = 1/2 ( R + Sqrt( R^2 + 4R)) or
r- = 1/2 ( R - Sqrt( R^2 + 4R))

and r^2, which is also equal to R(1+r) according to eq 1, is

r^2 = R/2 ( R+2 + Sqrt( R^2 + 4R))

or

r^2 = R/2 ( R+2 - Sqrt( R^2 + 4R))

Case 2) If D<0. This case happens if -4< R <0. Then the roots od eq 1 are complex numbers and r^2 too is a complex number. There are still 2 possible solutions.

For this case, the root of equation 1 are

r+ = 1/2 ( R + i Sqrt(- R^2 - 4R)) ;

r- = 1/2 ( R - i Sqrt( -R^2 - 4 R))

and you have as in the case 1 two solutions for r^2:

r^2 = R/2 ( R+2 +i Sqrt( -R^2 - 4R))

or

r^2 = R/2 ( R+2 - i Sqrt( -R^2 - 4R))

Answer 4

First, let's move everything that does not contain the r2 to the other side of the equal sign by subtracting 1/r1 from both sides.

1/R =1/r1+1/r2
-1/r1 - 1/r1

now, we're left with:
1/R - 1/r1 = 1/r2

Since 1 is divided by r2, we will need to undo that by multiplying both sides by r2.

r2 (1/R - 1/r1) = 1 is what we're left with.

Now, since r2 is multiplyied by (1/R - 1/r1) we're going to divde both sides by (1/R - 1/r1) to get the r2 alone.

r2 = 1/(1/R - 1/r1)

Answer 5

I would make the substitution u=1/r, then your equation becomes:

u^2 +u -1/R = 0

Solve via quadratic equation for u:

u=[-1 +/- Sqrt[1+4/R]/2

Remembering that u=1/r, your solutions for r are just the reciprocal of this expression above

Answer 6

r2=R(r1+1)

Answer 7

1/R = 1/r + 1/r^2

Let x =1/r

1/R = x + x^2
x^2 + x - 1/R = 0

x = -(-1 +/- sqrt(1 - 4/R))/2
= (1 +/- sqrt(1-4/R))/2

so
x^2 = (1 + (1-4/R) +/- 2 sqrt(1-4/R))/4
= (2 - 4/R +/- sqrt(1-4/R))/4

Thus r^2 = 1/x^2

r^2 = 4/(2-4/R +/- sqrt(1-4/R))

Answer 8

1/R = 1/r¹ + 1/r²

Multiply both sides of equation by r²R
r²R/R = r²R/r¹ + r²R/r²
r² = rR + R
r² - Rr = R

Complete the square
r² - Rr + (R/2)² = R + (R/2)²
(r-R/2)² = R + R²/4
r - R/2 = ±√(R + R²/4)
r = R/2 ± √[(4R+R²)/4]
= R/2 ± √(4R+R²)/2
r = R/2 ± √(4R+R²)/2

r₊ = R/2 + √(4R+R²)/2
r₋ = R/2 - √(4R+R²)/2

Answer 9

1/R = 1/r + 1/r2
1/r2 = 1/R - 1/r
(1/r2 ) x r4 = (1/R - 1/r) x r4
r4/r2 = r4/R - r4/r
r2 = r4/R - r3