QJEEM21Mab179: JEE Main 2021 Mathematics - Class 11 JEEM21Mab179 Solution: If , are roots of the equation , > and for each positive…

Question

Verified Solution • 2025–26 Session moderate

If

\alpha

,

\beta

are roots of the equation

{x^2} + 5(\sqrt 2 )x + 10 = 0

,

\alpha

>

\beta

and

{P_n} = {\alpha ^n} - {\beta ^n}

for each positive integer n, then the value of

\left( {{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}} \right)

is equal to _________.

Accepted Answer

{x^2} + 5\sqrt 2 x + 10 = 0

&

{P_n} = {\alpha ^n} - {\beta ^n}

(Given)

Now,

{{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}}

=

{{{{P_{17}}({P_{20}} + 5\sqrt 2 {P_{19}})} \over {{P_{18}}({P_{19}} + 5\sqrt 2 P_{18}^{})}}}

{{{P_{17}}({\alpha ^{20}} - {\beta ^{20}} + 5\sqrt 2 ({\alpha ^{19}} - {\beta ^{19}}))} \over {{P_{18}}({\alpha ^{19}} - {\beta ^{19}} + 5\sqrt 2 ({\alpha ^{18}} - {\beta ^{18}}))}}

{{{P_{17}}({\alpha ^{19}}(\alpha + 5\sqrt 2 ) - {\beta ^{19}}(\beta + 5\sqrt 2 ))} \over {{P_{18}}({\alpha ^{18}}(\alpha + 5\sqrt 2 ) - {\beta ^{18}}(\beta + 5\sqrt 2 ))}}

Since,

\alpha + 5\sqrt 2 = - 10/\alpha

..... (1)

&

\beta + 5\sqrt 2 = - 10/\beta

....... (2)

Now, put there values in above expression

= - {{10{P_{17}}{P_{18}}} \over { - 10{P_{18}}{P_{17}}}} = 1

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