Question

√1/1-t + √1-t/ = 13/6 find t​

Answer 1

for the given equation to be real,

t ≤ 1 or t ≤ 0

or √(1 - t) would be imaginary i.e. 1 - t < 0

$\sqrt{ \frac{1}{1 - t} } + \sqrt{1 - t} = \frac{13}{6} \\ \frac{1}{ \sqrt{1 - t} } + \sqrt{1 - t} = \frac{13}{6} \\ \frac{1 + ( \sqrt{1 -t })( \sqrt{1 - t} )}{ \sqrt{1 - t} } = \frac{13}{6 } \\ \frac{1 + 1 - t}{ \sqrt{1 - t} } = \frac{13}{6} \\ 2 - t = \frac{13}{6} \sqrt{1 - t} \\ {(2 - t)}^{2} = {( \frac{13}{6} \sqrt{1 - t} )}^{2} \\ 4 + {t}^{2} - 4t = \frac{169}{36} \times 1 - t \\ 36 {t}^{2} - 144t + 144 = 169 - 169t \\ 36 {t}^{2} - 144t + 169t + 144 - 169 = 0 \\ 36 {t}^{2} + 25t - 25 = 0 \\ 36 {t}^{2} + 45t - 20t - 25 = 0 \\ 9t(4t + 5) - 5(4t + 5) = 0 \\ (9t - 5)(4t + 5) = 0 \\ \\ 9t - 5 = 0 \\ or \\ 4t + 5 = 0 \\ \\ t = \frac{5}{9} \\ or \\ t = \frac{ - 5}{4} \\ \\ here \: t = \frac{5}{9} \: or \: \frac{ - 5}{4} satisfy \\ \\ t \leqslant 0 \: and \: t \leqslant 1$

hence, both the values are valid

Hence, t = 5/9 or -5/4