Question

If (p - x) : (q - x) be the duplicate ratio of p : q then show that: `1/"p" + 1/"q" = 1/"x"`

Answer 1

We have, 

`("p" - "x")/("q" - "x") = "p"^2/"q"^2`

⇒ q²(p - x) = p²(q - x)

⇒ pq2 - q2x = p2q - p2x

⇒ p2x - q2x = p2q - pq2

⇒ x(p2 - q2) =pq (p -q)

⇒ x(p - q)(p + q) = pq (p - q)

⇒ x = `"pq"/("p" + "q")`

⇒ `"pq"/("p" + "q") = 1/"x"`

⇒ `"p"/"pq" + "q"/"pq" = 1/"x"`

⇒ `1/"q" + 1/"p" = 1/"x"`

⇒ `1/"p" + 1/"q" = 1/"x"`

Answer 2

Given that (p − x) : (q − x) is the duplicate ratio of p : q, we need to prove that: 

`1/p + 1/q = 1/x`

The statement "duplicate ratio" means that the ratio of (p − x) to (q−x) is the square of the ratio p : q.

`(p-x)/(q-x) = (p/q)^2`

`p - x = (p/q)^2 (q-x)`

Expand the right side of the equation:

`p-x = p^2/q^2 (q-x)`

`p-x = p^2/q^2  q-p^2/q^2  x`

`p - x = p^2/q - p^2/q^2 x`

Now, let's collect the terms involving x on one side of the equation and the constant terms on the other side:

`p - p^2/q = x (p^2/q^2 -1)`

Now, factor the expression on the right-hand side:

`p - p^2/q = x ((p^2 - q^2)/q^2)`

Simplifying the left-hand side:

`p- p^2/q = (pq-p^2)/q`

`(pq-p^2)/q = x ((p^2-q^2)/q^2)`

Now, solve for x by multiplying both sides of the equation by q²

(q²) (pq − p²) = x(q) (p² − q²)

This simplifies to:

`x = (q(pq-p^2))/(p^2-q^2)`

At this point, we can conclude that the equation involves the values of p, q, and x, and the final goal is to demonstrate that:

`1/p + 1/q = 1/x`