# Let `f(x) = x^2+bx+c` and `g(x) = x^2+qx+r` , where `b, c, q, r in RR` and `c != r` Let `alpha, beta` be the roots of `g(x) = 0` It is given that `f(alpha)xxf(beta) = (c-r)^2 -(b-q) (cq-br)` prove...

Let `f(x) = x^2+bx+c` and `g(x) = x^2+qx+r` , where `b, c, q, r in RR` and `c != r` Let `alpha, beta` be the roots of `g(x) = 0`

It is given that `f(alpha)xxf(beta) = (c-r)^2 -(b-q) (cq-br)`

prove that if `f(x) = 0` and `g(x)=0` have a common root, then ` b-q, c-r ` and `cq - br` are in Geometric progression.

*print*Print*list*Cite

### 1 Answer

`g(x) = 0` has roots as alpha and beta.

It is given that `f(x) = 0` and `g(x) = 0` has a common root.

This means either alpha or beta is a root of `f(x) = 0` .

`f(alpha) = 0` or `f(beta) = 0.`

Therefore `f(alpha)xxf(beta) = 0`

`(c-r)^2-(b-q)(cq-br) = 0`

`(c-r)^2 = (b-q)(cq-br)`

`(c-r)/(b-q) = (cq-br)/(c-r)`

If `b-q,c-r` and `cq-br` follows a geometric progression there common ratios are equal.

That means `(c-r)/(b-q) = (cq-br)/(c-r)`

This relationship is obtained above.

*so `b-q,c-r` and `cq-br` follows a geometric progression*

**Sources:**