What is the inverse of (s^2-10s+21)/((s-3)^2(s+3)) ?

The inverse of (s^2-10s+21)/((s-3)^2(s+3)) is (1+sqrt(1-4s(-9s+7)))/(2s),(1-sqrt(1-4s(-9s+7)))/(2s)

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inverse of (s^2-10s+21)/((s-3)^2(s+3))

inverse

\frac{s ^{2} - 10 s + 21}{( s - 3 ) ^{2} ( s + 3 )}

\frac{1 + 1 - 4 s ( - 9 s + 7 )}{2 s}, \frac{1 - 1 - 4 s ( - 9 s + 7 )}{2 s}

Solution steps

y = \frac{s ^{2} - 10 s + 21}{( s - 3 ) ^{2} ( s + 3 )}

Replace

s

with

y

s = \frac{y ^{2} - 10 y + 21}{( y - 3 ) ^{2} ( y + 3 )}

Solve for

y, s = \frac{y ^{2} - 10 y + 21}{( y - 3 ) ^{2} ( y + 3 )}

\frac{1 + 1 - 4 s ( - 9 s + 7 )}{2 s}, \frac{1 - 1 - 4 s ( - 9 s + 7 )}{2 s}

What is the inverse of (s^2-10s+21)/((s-3)^2(s+3)) ?
The inverse of (s^2-10s+21)/((s-3)^2(s+3)) is (1+sqrt(1-4s(-9s+7)))/(2s),(1-sqrt(1-4s(-9s+7)))/(2s)