If x4−79x2+1=0x^4-79x^2+1=0x4−79x2+1=0then the value of x+x−1x+x^{-1}x+x−1 can be; A9B5C7D8Answer: A. 9 Read Explanation: x4−79x2+1=0x^4-79x^2+1=0x4−79x2+1=0divide the equation with x2x^2x2x4/x2−79x2/x2+1/x2=0x^4/x^2-79x^2/x^2+1/x^2=0x4/x2−79x2/x2+1/x2=0x2−79+1/x2=0x^2-79+1/x^2=0x2−79+1/x2=0x2+1/x2=79x^2+1/x^2=79x2+1/x2=79we have ,If x+1/x=kx +1/x=kx+1/x=kthenx2+1/x2=k2−2x^2+1/x^2=k^2-2x2+1/x2=k2−2So here,k2−2=79k^2-2=79k2−2=79k2=81k^2=81k2=81k=9k=9k=9x+x−1=x+1/x=9x+x^{-1}=x+1/x=9x+x−1=x+1/x=9 Read more in App