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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??

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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