Quadratic Equation: ratio of roots

[InguteInga]

Hi guys,

I am struggling with the question below. Any help would be very much appreciated..

If the roots of the equation px^2 + rx + r = 0 are in the ratio a:b, then the value of √(b/a) + √(a/b) is:

A: √(1/r)
B: -√(r/p)
C: √(r/p)
D: √(1/p)
E: √(3/p)

Many thanks!

[puneetkhurana2000]

√(b/a) + √(a/b) = (a+b)/√ab ....(1) by taking LCM as √ab

Now sum of the roots of the equation ax^2 + bx + c = 0 is -b/a and product of roots is c/a.

So, px^2 + rx + r = 0 has sum of roots as -r/p and product as r/p.

Roots can be expressed as ak and bk ....where k is a constant (as a:b is the ratio of roots and not the roots)

Plugging the values we get a*k+b*k = -r/p and a*k*b*k = r/p
Further solving we get (a+b)k = -r/p and ab*k^2 = r/p .....(2)

Plugging the values from equation (2) of (a+b) and ab in (1) we get (a+b)/√ab = -√(r/p

Hope this helps!!!

[InguteInga]

√(b/a) + √(a/b) = (a+b)/√ab ....(1) by taking LCM as √ab

Now sum of the roots of the equation ax^2 + bx + c = 0 is -b/a and product of roots is c/a.

So, px^2 + rx + r = 0 has sum of roots as -r/p and product as r/p.

Roots can be expressed as ak and bk ....where k is a constant (as a:b is the ratio of roots and not the roots)

Plugging the values we get a*k+b*k = -r/p and a*k*b*k = r/p
Further solving we get (a+b)k = -r/p and ab*k^2 = r/p .....(2)

Plugging the values from equation (2) of (a+b) and ab in (1) we get (a+b)/√ab = -√(r/p

Hope this helps!!!

puneetkhurana2000,

Hi! Thank you, yes, it is very helpful indeed.

But could you please elaborate on plugging values from (2) equation into (1) as I am left with those k coefficients.. I am sorry if this is very straight forward, but I am still getting confused.

Many thanks once again!

[puneetkhurana2000]

(a+b)/√ab = Multiply the num and den by k ... (a+b)k/k√ab = (a+b)*k/√ab*k^2

Now from (2) we know (a+b)*k and ab*k^2

Plugging we get -(r/p)/√(r/p) = -√(r/p)

Hope this helps!!!