If `a\ a n d\ b` are the roots of `x^2-3x+p=0\ a n d\ c ,\ d` are the roots `x^2-12 x+q=0` where `a , b , c , d` form a G.P. Prove that `(q+p):(q-p)=17 : 15.`

Answer

N/a

Explanation

`because` a and b are the roots of equation
`x^(2) - 3x + p = 0`
`therefore` a + b = 3 and ab = p
Similarly, c and d are the roots of equation
`x^(2) - 3x + p = 0`
`therefore` c + d = 12 and cd = q
Then q + p = cd + ab and q - p = cd - ab
`((q +p))/((q - p)) = (cd + ab)/(cd - ab)` ...(1)
`because` a, b, c, d are in G.P.
`therefore` `(a)/(b) = (b)/(c) = (c)/(b)`
`therefore` `(a)/(b) + 1 = (b)/(c) + 1 = (c)/(d) + 1`
`therefore` `(a + b)/(b) = (b + c)/(c) = (c + d)/(d)`
`therefore` `(3)/(b) = (b + c)/(c) = (12)/(d) rArr (1)/(b) = (4)/(d)`
`rArr` d = 4b
`because` a + b = 3 and c + d = 12 `rArr` 4a + 4b = c + d
and 4b = d `rArr` c = 4a
`therefore` c = 4a and d = 4b `rArr` cd = 16ab
Put the value of cd in equation (1)
`(q + p)/(q - p) = (16ab + ab)/(16ab - ab) rArr (q + p)/(q - p) = (17ab)/(15ab)`
`rArr` `(q + p)/(q - p) = (17)/(15)`
So, (q + p) : (q - p) = 17 : 15 Hence Proved.