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Miscellaneous Exercise · Q9

Q.If aa and bb are the roots of x2−3x+p=0x^2 - 3x + p = 0 and cc, dd are roots of x2−12x+q=0x^2 - 12x + q = 0, where aa, bb, cc, dd form a G.P. Prove that (q+p):(q−p)=17:15(q + p) : (q - p) = 17 : 15.

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When four numbers in G.P. are roots of two quadratics, their common ratio and the relationship between sum/product of roots forces a unique ratio between the constants. We find (q+p):(q−p)=17:15(q + p) : (q - p) = 17 : 15.

The heart of this problem lies in connecting two pieces of algebra: Vieta's formulas (which relate roots to coefficients) and the structure of a geometric progression. When four terms form a G.P., each term is the previous one multiplied by a common ratio rr. This rigid structure, combined with the constraints from the two quadratics, will lock down the relationship between pp and qq.

Let the four terms of the G.P. be aa, arar, ar2ar^2, ar3ar^3 where rr is the common ratio. We need to assign these to the roots of our two equations.

Setting up the correspondence

Since aa and bb are roots of the first equation and cc, dd are roots of the second, and all four form a G.P., we have:

  • a,b,c,da, b, c, d in G.P.

The natural assignment is a=aa = a, b=arb = ar, c=ar2c = ar^2, d=ar3d = ar^3.

Applying Vieta's formulas

  1. For the first quadratic x2−3x+p=0x^2 - 3x + p = 0 with roots aa and arar:

    Sum of roots: a+ar=3a + ar = 3, so a(1+r)=3a(1 + r) = 3.

    Product of roots: a⋅ar=pa \cdot ar = p, so a2r=pa^2 r = p.

  2. For the second quadratic x2−12x+q=0x^2 - 12x + q = 0 with roots ar2ar^2 and ar3ar^3:

    Sum of roots: ar2+ar3=12ar^2 + ar^3 = 12, so ar2(1+r)=12ar^2(1 + r) = 12.

    Product of roots: ar2⋅ar3=qar^2 \cdot ar^3 = q, so a2r5=qa^2 r^5 = q.

Finding the common ratio …

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