NTSE Stage 2 SAT 2019 Exam Question Paper

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Question : 46 of 100

Marks: +1, -0

If

x={3+√5}/2

and

y=x^3

,then y satisfies the quadratic equation

$y^{2}-18y+1=0$
$y^{2}+18y+1=0$
$y^{2}-18y-1=0$
$y^{2}+18y-1=0$

Solution:

x=\;\;{3+√{5}}/{2}

⇒x^{3}=(\;\;{3+√{5}}/{2})^{3}=\;\;{27+5√{5}+9√{5}(3+√{5})}/{8}=9+4√{5}

⇒y=x^{3}=9+4√{5}

⇒ One root is

9+4√5

∴ Other is root

9-4√5

∴ Sum of roots

=9+4√{5}+9-4√{5}=18

Product of roots

=(9+4√{5})(9-4√{5})=1

∴ Required equation is:

y^2-18y+1=0

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