# In this equation: t = r * sqrt((n - 2)/(1 - r^2)) can I solve for r?

### 8 Answers

- 1 month ago
t = r * sqrt((n - 2)/(1 - r^2)) solve for r

r / sqr ( 1- r ² ) = t / sqr(n - 2) squaring both sides

r ² / ( 1 - r ² ) = t ² / ( n - 2 )

r ² = t ² / ( n - 2 ) * ( 1 - r ² )

r ² = t ² / ( n - 2 ) - t ² / ( n - 2 ) r ²

r ² + t ² / ( n - 2 ) r ² = t ² / ( n - 2 ) factor r ²

( 1 + t ² / ( n - 2 ) ) * r ²

= t ² / ( n - 2 )

r ² = t ² / ( n - 2 ) / ( 1 + t ² / ( n - 2 ) )

r = sqr ( t ² / ( n - 2 ) / ( 1 + t ² / ( n - 2 ) ) )

- PhilipLv 61 month ago
t = r*sqrt[(n-2)/(1-r^2)].;

Square both sides getting t^2 = r^2(n-2)/(1-r^2). Multiply both sides by (1-r^2)

getting t^2(1-r^2) = (n-2)r^2. Add t^2*r^2 to both sides getting;

t^2 = (t^2+n-2)r^2. Divide both sides by (t^2+n-2) getting;

[t^2/(t^2+n-2)] = r^2. Take square root of both sides getting;

r = (+/-)sqrt[t^2/(t^2+n-2)].

- PinkgreenLv 71 month ago
t=rsqr[(n-2)/(1-r^2)]

=>

(t/r)^2=(n-2)/(1-r^2)

=>

(1-r^2)/r^2=(n-2)/t^2

=>

1/r^2=(n-2)/t^2+1

=>

1/r^2=(t^2+n-2)/t^2

=>

r^2=t^2/(t^2+n-2)

=>

r= t/sqr(t^2+n-2), take +ve r only.

- TomVLv 71 month ago
t = r√((n-2)/(1-r²))

Square both side to clear the radical sign. Note, however, this operation has the potential of introducing extraneous solutions, so the results will have to be verified against the original equation to filter out any such solutions

t² = r²(n-2)/(1-r²)

Now, use whatever algebraic manipulations are necessary to isolate r on one side of the equation.

t²(1-r²) = r²(n-2)

t² - t²r² = r²(n-2)

t² = r²(n-2) + t²r²

r²(n-2+t²) = t²

r² = t²/(t²+n-2)

Here is where it gets tricky. If you simply take the square root of both sides, the possibility of extraneous solutions rears its ugly head.

r = ±√[t²/(t²+n-2)] = ±t/√(t²+n-2)

The ± sign implies two solutions, but only one of the solutions is valid. From the original equation we see that t and r must have the same sign. If t > 0, then r > 0 and if t < 0, then r < 0

The correct solution is as follows:

r = t/√(t² + n - 2)

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- llafferLv 71 month ago
Is that:

t = r√[(n - 2) / (1 - r²)]

To solve this for r, I'd start with squaring both sides to get rid of the radical:

t² = r²(n - 2) / (1 - r²)

Now multiply both sides by the denominator to get rid of the fractions:

t²(1 - r²) = r²(n - 2)

We already have r² factored out of the right half, so simplify the left half of the equation:

t² - r²t² = r²(n - 2)

Move the r² term to the right side:

t² = r²(n - 2) + r²t²

Factor out r²:

t² = r²(n - 2 + t²)

Divide both sides by the trinomial on the right side:

t² / (n - 2 + t²) = r²

I'll put the trinomial into polynomial order at this point, then get the square root of both sides:

t² / (t² + n - 2) = r²

± √[t² / (t² + n - 2)] = r

- PopeLv 71 month ago
t = r√[(n - 2)/(1 - r²)]

t² = r²(n - 2)/(1 - r²)

t²(1 - r²) = r²(n - 2)

t² - t²r² = r²(n - 2)

t² = r²(n - 2 + t²)

r² = t²/(n - 2 + t²)

r = t/√(n - 2 + t²)

It is tempting to write ± there, but notice in the given equation that r and t must have the same sign.

- UserLv 71 month ago
Yes

t/r = sqrt (etc)

(t/r)^2 = (n-2)/(1-r^2)

t^2 / r^2 = (n-2)/(1-r^2)

t^2 (1 - r^2) = (n - 2) r^2

t^2 - r^2t^2 = nr^2 - 2r^2

t^2 = nr^2 - 2r^2 + 2t^2r^2

t^2 = r^2(n - 2 + 2t^2)

t^2/(n - 2 + 2t^2) = r^2

r = +/- sqrt [t^2/(n - 2 + 2t^2)]

- billrussell42Lv 71 month ago
t = r√((n - 2)/(1 - r²))

t² = r²((n - 2)/(1 - r²))

t²(1 - r²) = r²(n - 2)

t² – t²r² = r²(n - 2)

t²r² – r²(n - 2) = t²

r²(t² – n + 2) = t²

r² = t² / (t² – n + 2)

r = √(t² / (t² – n + 2))

r = t√(1 / (t² – n + 2))