# if x^2-px+q=0 & x^2-ax+b=0 have one common root and other root of 2nd is reciprocal of other root of first.then prove (q-b)^2=(bq)(p-a)^2

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You need to come up with the substitutions: `x_1,x_2` for the roots of equation `x^2-px+q=0` and `x_1, 1/x_2` for the roots of the equation `x^2-ax+b=0` .

You should use Vieta's relations such that:

`x_1 + x_2 = p`

`x_1*x_2 = q`

`x_1 + 1/x_2 = a =gt x_1*x_2 + 1 = a*x_2`

`q + 1 = a*x_2 =gt x_2 = (q+1)/a`

`x_1 = p - (q+1)/a`

`x_1/x_2 = b =gt x_1/(q/x_1) = b`

`(x_1)^2 = bq`

If you raise to square `x_1 + x_2` yields:

`(x_1 + x_2)^2=x_1^2 + 2x_1*x_2 + x_2^2 = p^2`

`bq + 2q + x_2^2 = p^2`

`(x_1 + 1/x_2)^2 = x_1^2 + 2(x_1/x_2) + (1/x_2)^2`

`(x_1 + 1/x_2)^2 = bq + 2b + (1/x_2)^2 = a^2`

If you subtract`bq + 2b + (1/x_2)^2 = a^2 ` from `bq + 2q + x_2^2 = p^2` yields:

`bq + 2b + (1/x_2)^2 - bq- 2q- x_2^2 = a^2 - p^2`

`2(b-q) + (1/x_2)^2 - x_2^2 = a^2 - p^2`

You need to substitute `x_2 = (q+1)/a` such that:

`2(b-q) + a^2/((q+1)^2) - ((q+1)^2)/a^2 = a^2 - p^2`

**`2(b-q) + (a/(q+1) - (q+1)/a)(a/(q+1) +(q+1)/a) = a^2 - p^2` **