Problem: $\sqrt{x}+\frac{1}{\sqrt{x}}$ = a then field the valu of $x^{2}+\frac{1}{x^{2}}$
View all: SHAJALAL ISLAMI BANK(TSO) WRITTEN QUESTION (MATH) SOLVE | 2016
Correct Answer: The value is a$^{4}$- 4a$^{2}$+ 2.
Explanation:
Given $\sqrt{x}+\frac{1}{\sqrt{x}}$
=> ( $\sqrt{x}+\frac{1}{\sqrt{x}}$ )$^{2}$ = a$^{2}$
=> $(\sqrt{x})^{2}$+2$\times$$\sqrt{x}$$\times$$\frac{1}{\sqrt{x}}$+$(\frac{1}{\sqrt{x}})$$^{2}$ = a$^{2}$ [(a+b)$^{2}$ = a$^{2}$+2ab+b$^{2}$]
=> x+2+$\frac{1}{x}$ = a$^{2}$
=> x+$\frac{1}{x}$ = a$^{2}$-2
=>(x+$\frac{1}{x}$)$^{2}$ = (a$^{2}$-2)$^{2}$
=> x$^{2}$+2$\times$x$\times$+$\frac{1}{x}$+($\frac{1}{x}$)$^{2}$ = (a$^{2}$)$^{2}$-2$\times$a$^{2}$$\times$2+2$^{2}$
=> x$^{2}$+2+$\frac{1}{x^{2}}$ = a$^{4}$-4a$^{2}$+4
=> x$^{2}$+$\frac{1}{x^{2}}$ = a$^{4}$-4a$^{2}$+4-2
x$^{2}$+$\frac{1}{x^{2}}$ = a$^{4}$-4a$^{2}$+2
