1.

Solved the equation by using quadratic formula `a(x^(2)+1)=(a^(2)+1)x,ane0`.

Answer» Given equation is
`a(x^(2)+1)=(a^(2)+1)x`
`impliesax^(2)+a=a^(2)x+x`
`impliesax^(2)-x(a^(2)+1)+a=0`
Comparing with `Ax^(2)+Bx+C=0`, we get `A=a,B=-(a^(2)+1)andC=a`
`:.x=(-B+-sqrt(B^(2)-4AC))/(2A)`
`impliesx=(-[-(a^(2)+1)]+-sqrt({-(a^(2)+1)}^(2)-4xxaxxa))/(2a)`
`impliesx=((a^(2)+1)+-sqrt(a^(4)+1+2a^(2)-4a^(2)))/(2a)`
`impliesx=((a^(2)+1)+-sqrt((a^(2)-1))^(2))/(2a)`
`impliesx=((a^(2)+1)+-(a^(2)-1))/(2a)`
`impliesx=(a^(2)+1+a^(2)-1)/(2a)and(a^(2)+1-a^(2)+1)/(2a)`
`impliesx=(2a^(2))/(2a)and(2)/(2a)`
`impliesx=aand(1)/(a)` are roots of the equation.


Discussion

No Comment Found

Related InterviewSolutions