Question

What is the approach of this question?

Answer 1

1) `a^2/p^2`

Explanation:

This is my first attempt and may be more complicated than necessary, but:

Try keeping the problem fairly symmetric...

Let `m` be the mean of `alpha, beta, gamma, delta` and `h` half of the common difference.

Then:

`{ (alpha = m - 3h), (beta = m-h), (gamma = m+h), (delta = m+3h) :}`

and:

`ax^2+bx+c = a(x-alpha)(x-beta)`

`color(white)(ax^2+bx+c) = a(x-m+3h)(x-m+h)`

`color(white)(ax^2+bx+c) = ax^2-2(m-2h)ax+(m^2-4hm+3h^2)a`

So:

`{ (b = -2(m-2h)a), (c = m^2-4hm+3h^2) :}`

and:

`D_1 = b^2-4ac`

`color(white)(D_1) = 4a^2((m-2h)^2-(m^2-4hm+3h^2))`

`color(white)(D_1) = 4a^2((m^2-4hm+4h^2)-(m^2-4hm+3h^2))`

`color(white)(D_1) = 4a^2h^2`

We can then simply replace `h` with `-h` and `a` with `p` to find:

`D_2 = 4p^2h^2`

So:

`D_1/D_2 = (4a^2h^2)/(4p^2h^2) = a^2/p^2`

Answer 2

1) `a^2/p^2`

Explanation:

Here's a simpler method...

`ax^2+bx+c = a(x-alpha)(x-beta)`

`color(white)(ax^2+bx+c) = a(x^2-(alpha+beta)x+alphabeta)`

`color(white)(ax^2+bx+c) = ax^2-(alpha+beta)ax+alphabetaa`

So:

`D_1 = b^2-4ac`

`color(white)(D_1) = a^2((alpha+beta)^2-4alphabeta)`

`color(white)(D_1) = a^2(alpha^2+2alphabeta+beta^2-4alphabeta)`

`color(white)(D_1) = a^2(alpha^2-2alphabeta+beta^2)`

`color(white)(D_1) = a^2(alpha-beta)^2`

Similarly:

`D_2 = p^2(gamma-delta)^2`

But `alpha, beta, gamma, delta` are in arithmetic progression. So:

`gamma-delta = beta-alpha`

and:

`D_1/D_2 = (a^2(alpha-beta)^2)/(p^2(gamma-delta)^2) = a^2/p^2`