Question

Can you tell me how x^(-1/2)=t-3 become x=t^2-6t+9

Answer 1

For what you need

x have power 1/2 (not -1/2)

because x^(1/2) = √x

but x^(-1/2) = 1/√x

for given condition

x^(1/2) = t - 3

√x=t-3

squaring both side

x = (t-3)^2

x = t^2 -6t + 9

or

for x^(-1/2)

x^(-1/2) = t -3

1/√x = t -3

squaring both side

1/x = t^2 -6t +9

so now you understand it

ok

Answer 2

{x}^{-½}= t-3

squaring both sides, we get

→ ({x}^{-½})² = (t-3)²

→ {x}^{-1} = t² -2(t)(3) + 9

→ 1/x = t² -6t +9

Here if {x}^{½} is taken then

x = t² -6t +9

Hope it helps!!