Question

If alpha (plus minus) sqrt beta be the roots of the equation x^2+px+q=0,prove that 1/alpha (plus minus) 1/sqrt beta will be the roots of the equation (p^2-4q) (p^2x^2+4px)-16q=0?

Answer 1

See the proof below.

Explanation:

`x=a+-sqrtb`
`x_1=a-sqrtb,x_2=a+sqrtb`
`x^2+px+q=0`
Sum of roots:
`a-sqrtb+a+sqrtb=-p/1`
`p=-2a`
Product of roots;
`(a-sqrtb)(a+sqrtb)=q/1`
`q=a^2-b`

`(p^2-4q)(p^2x^2+4px)-16q=0`
`(4a^2-4a^2+4b)(4a^2x^2-8ax)-16a^2+16b=0`
`16a^2bx^2-32abx-16a^2+16b=0`
`ab(ax^2-2x)-a^2+b=0`
`ax^2-2x=(a^2-b)/(ab)`
`a(x^2-(2/a)x+(1/a)^2)=(a^2-b)/(ab)+1/a`
`(x-1/a)^2=(a^2-b)/(a^2b)+1/a^2`
`(x-1/a)^2=(a^2-b+b)/(a^2b)`
`(x-1/a)^2=1/b`
`x-1/a=+-1/sqrtb`
`x=1/a+-1/sqrtb`
Hence proved.