CandyFloss~

1 .

THE ANS IS

x = \frac{a^2 + b^2 }{a+ b}

posting the complete solution

on slight rearrangement

\frac{x - a}{b} - \frac{b}{x- a} = \frac{a}{x- b }-\frac{x-b}{a}

\frac{\left(x- a \right)^ 2 - (b^2) }{b \left(x- a \right)} = \frac{a^ 2 - \left(x-b \right)^ 2}{( x- b). a}

again rearranging we get

\frac{\left(x- a \right)^ 2 - (b^2) }{(a^ 2) - \left(x- b \right)^2 } = \frac{b (x-a)}{a( x- b). }

now use a^ 2 - b^2 pharmoola

\frac{\left(x- a+ b \right)\left(x- a- b \right)}{\left(a+ x- b \right)\left(a-x + b \right)} = \frac{\left(x-a + b \right)(x - a- b )}{\left(a + x - b \right)\left(-( -a + x- b) \right)}

\frac{ x- a + b }{-a - x+ b } = \frac{b\left(x-a \right)}{a\left(x-b \right)}

now on rearranging we get

( x- a + b )a ( x- b) = b ( x- a ) .( b- a - x )

simplifying both sides

( ax - a^2 + ab ) ( x- b) = ax^2 - axb - a^ 2x + a^2 b + abx - ab^2

---1

( b^2 - ba - bx ) ( x- a ) = b^ 2x - b^2a - bax + ba^2 - bx^2 + bax---2

equate 1 and 2

ax^ 2 - a^ 2 x + a^2 b - ab^2 = b^2 x - bx^2 + a^2b - ab^2

ax^ 2 - a^ 2 x = b^2 x - bx^2

ax - a^2 = b^2 - bx

ax + bx = a^2 + b^2

x ( a + b ) = a^2 + b^2

x =\frac{a^2 + b^2 }{ a+ b}

no need to do so lengthy
on shifting terms from lhs to rhs and rhs to lhs we hav
\frac{x-a}{b}-\frac{a}{x-b}=\frac{b}{x-a}-\frac{x-b}{a}
now siplify seperately rhs and lhs
u get the ans
now that thing was quiet obvious
my q is since it comes out to be a function of 3 degree will der be any more roots bec in this we r cancelling numerator having x

\\\texttt{Q-4 } \\\texttt{we have dicriminant > 0 } \\ a > b \\ \\\texttt{let S(n) be the left hand site of inequality } \\ \\ S(n) < \lim_{n\rightarrow \infty}S(n)=\frac{(\alpha + \beta )^2}{(\alpha -\beta )^2}

other roots are

x=0 x= a+b

equating

(x-a)(x-b)-ab=0

by simple observation we can see that x= o is a root since it satisfies

equation