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suppose x1, x2 be the roots of ax^2+bx+c=0 and x^3,x^4 be the roots of px^2+qx+r=0. (i) x1, x2 , 1/x3, 1/x4 are in AP then (b^2 -4ac)/(q^2 -4pr) is equal to??? (ii) If a,b,c are in GP as well as x1,x2,x3,x4 are in GP, then p,q,r are in?? (iii)If x1,x2,x3,x4 are in GP, then its common ratio is??

8 years ago

Answers : (1)

abc
12 Points
							
Kindly confirm on the roots of the second equation. I suppose it must be x₃, x₄ and not x³, x⁴ as mentioned. 
On this supposition, let me provide the solution for the first question. If it is OK, kindly give a feed back, basing on which let me give the solutions for the rest of questions. 

[Even it is not correct, kindly give me a feed back, so that I can delete my solution - there by avoiding unnecessary TD ratings. Thanks and kind of your feed back] 

Solution for question 1. 

i) Applying properties of roots of Q.E., 
x₁ + x₂ = -b/a and x₁*x₂ = c/a --------- (1) 
x₃ + x₄ = -q/p and x₃*x₄ = r/p ----------- (2) 

ii) Since, x₁, x₂, 1/x₃, 1/x₄ are in A.P., 
x₂ - x₁ = 1/x₄ - 1/x₃ [Definition of AP - Difference of successive terms remain same] 

==> x₂ - x₁ = (x₃ - x₄)/(x₃*x₄) 

Squaring both sides, (x₂ - x₁)² = (x₃ - x₄)²/(x₃*x₄)² 

Applying algebraic identity, (a - b)² = (a + b)² - 4ab to the above, 
(x₂ + x₁)² - 4x₁*x₂ = {(x₃ + x₄)² - 4x₃*x₄}/(x₃*x₄)² 

Substituting the values from equations (1) & (2) to the above, 
(b²/a²) - 4c/a = [(q²/p²) - 4r/p]/(r/p)² 

This simplifies as: (b² - 4ac)/a² = (q² - 4pr)/r² 

So, (b² - 4ac)/(q² - 4pr) = (a/r)²
6 years ago
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