SSC CHSL 12 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

© examsiri.com

Question : 73 of 99

Marks: +1, -0

If

x^{2}+(4-√{3}) x-1=0

, then what is the value of

x^{2}+\;{1}/{x^{2}} ?

$9-8 √{3}$
$21-12 √{3}$
$21-8 √{3}$
$17-8 √{3}$

Solution:

Given,

x^{2}+(4-√{3}) x-1=0

⇒ \;\; x^{2}-1=(-4+√{3}) x

⇒ \;\; \;{x^{2}-1}/{x}=√{3}-4

⇒ \;\; x-\;{1}/{x}=√{3}-4

We know that,

(a-b)^{2} \;\; =a^{2}+b^{2}-2 a b

∴ (x-\;{1}/{x})^{2} \;\; =x^{2}+\;{1}/{x^{2}}-2 × x × \;{1}/{x}

∴ x^{2}+\;{1}/{x^{2}} \;\; =(√{3}-4)^{2}+2

\;\; =3+16-8 √{3}+2

\;\; =21-8 √{3}

© examsiri.com

Go to Question:

Prev Question

Next Question