Given quadratic in terms of roots α , β \alpha,\beta α , β
The equation is
x 2 − ( 5 + 3 log 3 5 − 5 log 5 3 ) x + 3 ( 3 ( log 3 5 ) 1 3 − 5 ( log 5 3 ) 2 3 − 1 ) = 0. x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0. x 2 − ( 5 + 3 l o g 3 5 − 5 l o g 5 3 ) x + 3 ( 3 ( l o g 3 5 ) 3 1 − 5 ( l o g 5 3 ) 3 2 − 1 ) = 0. So,
α + β = 5 + 3 log 3 5 − 5 log 5 3 , \alpha+\beta=5+3^{\sqrt{\log _3 5}}-5^{\sqrt{\log _5 3}}, α + β = 5 + 3 l o g 3 5 − 5 l o g 5 3 , and
α β = 3 ( 3 ( log 3 5 ) 1 / 3 − 5 ( log 5 3 ) 2 / 3 − 1 ) . \alpha\beta=3\left(3^{(\log_3 5)^{1/3}}-5^{(\log_5 3)^{2/3}}-1\right). α β = 3 ( 3 ( l o g 3 5 ) 1/3 − 5 ( l o g 5 3 ) 2/3 − 1 ) .
Simplify the strange exponential terms
Let
a = log 3 5. a=\log_3 5. a = log 3 5. Then
log 5 3 = 1 a . \log_5 3=\frac{1}{a}. log 5 3 = a 1 . Now simplify:
(i) 3 log 3 5 3^{\sqrt{\log_3 5}} 3 l o g 3 5
3 a . 3^{\sqrt{a}}. 3 a . (ii) 5 log 5 3 5^{\sqrt{\log_5 3}} 5 l o g 5 3
Since 5 = 3 a 5=3^a 5 = 3 a
5 1 / a = ( 3 a ) 1 / a = 3 a / a = 3 a . 5^{\sqrt{1/a}}=(3^a)^{1/\sqrt a}=3^{a/\sqrt a}=3^{\sqrt a}. 5 1/ a = ( 3 a ) 1/ a = 3 a / a = 3 a . Hence,
3 log 3 5 = 5 log 5 3 . 3^{\sqrt{\log_3 5}}=5^{\sqrt{\log_5 3}}. 3 l o g 3 5 = 5 l o g 5 3 . Therefore,
α + β = 5. \alpha+\beta=5. α + β = 5.
Simplify the product α β \alpha\beta α β
We need to simplify
3 ( log 3 5 ) 1 / 3 − 5 ( log 5 3 ) 2 / 3 . 3^{(\log_3 5)^{1/3}}-5^{(\log_5 3)^{2/3}}. 3 ( l o g 3 5 ) 1/3 − 5 ( l o g 5 3 ) 2/3 . Again let a = log 3 5 a=\log_3 5 a = log 3 5 5 = 3 a 5=3^a 5 = 3 a log 5 3 = 1 / a \log_5 3=1/a log 5 3 = 1/ a
Then
5 ( log 5 3 ) 2 / 3 = ( 3 a ) ( 1 / a ) 2 / 3 = 3 a ⋅ a − 2 / 3 = 3 a 1 / 3 . 5^{(\log_5 3)^{2/3}}=(3^a)^{(1/a)^{2/3}}=3^{a\cdot a^{-2/3}}=3^{a^{1/3}}. 5 ( l o g 5 3 ) 2/3 = ( 3 a ) ( 1/ a ) 2/3 = 3 a ⋅ a − 2/3 = 3 a 1/3 . Also,
3 ( log 3 5 ) 1 / 3 = 3 a 1 / 3 . 3^{(\log_3 5)^{1/3}}=3^{a^{1/3}}. 3 ( l o g 3 5 ) 1/3 = 3 a 1/3 . Thus these two are equal, so
3 ( log 3 5 ) 1 / 3 − 5 ( log 5 3 ) 2 / 3 = 0. 3^{(\log_3 5)^{1/3}}-5^{(\log_5 3)^{2/3}}=0. 3 ( l o g 3 5 ) 1/3 − 5 ( l o g 5 3 ) 2/3 = 0. Hence,
α β = 3 ( 0 − 1 ) = − 3. \alpha\beta=3(0-1)=-3. α β = 3 ( 0 − 1 ) = − 3. So we have:
α + β = 5 , α β = − 3. \alpha+\beta=5,\qquad \alpha\beta=-3. α + β = 5 , α β = − 3.
Find sum and product of new roots
The new roots are
α + 1 β , β + 1 α . \alpha+\frac{1}{\beta},\qquad \beta+\frac{1}{\alpha}. α + β 1 , β + α 1 . Let these be r 1 r_1 r 1 r 2 r_2 r 2
Sum:
r 1 + r 2 = α + β + 1 α + 1 β r_1+r_2=\alpha+\beta+\frac{1}{\alpha}+\frac{1}{\beta} r 1 + r 2 = α + β + α 1 + β 1 = ( α + β ) + α + β α β = (\alpha+\beta)+\frac{\alpha+\beta}{\alpha\beta} = ( α + β ) + α β α + β = 5 + 5 − 3 = 5 − 5 3 = 10 3 . =5+\frac{5}{-3}=5-\frac53=\frac{10}{3}. = 5 + − 3 5 = 5 − 3 5 = 3 10 . Product:
r 1 r 2 = ( α + 1 β ) ( β + 1 α ) . r_1r_2=\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right). r 1 r 2 = ( α + β 1 ) ( β + α 1 ) . Expand:
= α β + 1 + 1 + 1 α β =\alpha\beta+1+1+\frac{1}{\alpha\beta} = α β + 1 + 1 + α β 1 = α β + 2 + 1 α β . =\alpha\beta+2+\frac{1}{\alpha\beta}. = α β + 2 + α β 1 . Using α β = − 3 \alpha\beta=-3 α β = − 3
r 1 r 2 = − 3 + 2 − 1 3 = − 1 − 1 3 = − 4 3 . r_1r_2=-3+2-\frac13=-1-\frac13=-\frac43. r 1 r 2 = − 3 + 2 − 3 1 = − 1 − 3 1 = − 3 4 .
Form the required quadratic
A quadratic with roots r 1 , r 2 r_1,r_2 r 1 , r 2
x 2 − ( r 1 + r 2 ) x + r 1 r 2 = 0. x^2-(r_1+r_2)x+r_1r_2=0. x 2 − ( r 1 + r 2 ) x + r 1 r 2 = 0. So,
x 2 − 10 3 x − 4 3 = 0. x^2-\frac{10}{3}x-\frac43=0. x 2 − 3 10 x − 3 4 = 0. Multiplying by 3 3 3
3 x 2 − 10 x − 4 = 0. 3x^2-10x-4=0. 3 x 2 − 10 x − 4 = 0.
Match with options
This is Option B .
3 x 2 − 10 x − 4 = 0 \boxed{3x^2-10x-4=0} 3 x 2 − 10 x − 4 = 0