1.

Given that `alpha` and `beta` are the roots of the equation `x^2=x+7` Prove that `(a) 1/alpha=(alpha-1)/7` `(b) alpha^3=8 alpha+7` Find the numerical value of `alpha/beta+beta/alpha`

Answer» Given equation is, `x^2 = x+7`
`=>x^2-x-7 = 0`
(a) As `alpha ` is root of this equation, `x = alpha` should satisfy this equation.
`alpha^2-alpha-7 = 0`
`=>alpha(alpha-1) = 7`
`=>1/alpha = (alpha-1)/7`

(b) `alpha^2-alpha-7 = 0`
`=>alpha^2 = alpha+7`
`=>alpha^3 = alpha^2+7alpha `
`=>alpha^3 = (alpha+7)+7alpha`
`=>alpha^3 = 8alpha+7`

As, `alpha and beta` are the roots of this equation,
`:. alpha+beta = -b/a = 1`
`alphabeta = c/a = -7`
Now, `alpha/beta+beta/alpha = (alpha^2+beta^2)/(alphabeta)`
`=((alpha+beta)^2-2alphabeta)/(alphabeta)`
`=(1^2-2(-7))/(-7) = -15/7`
`:. alpha/beta+beta/alpha = -15/7`


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