Hey guys plz solve this
Answers
Given : E₁ = ax² + bx + c , E₂ = bx² + cx + a , E₃ = cx² + bx + a
To find : Quadratic expression having zeroes that are common to E₂ & E₃ but not to E₁
Solution:
E₁ = ax² + bx + c Zeroes P & R
E₂ = bx² + cx + a Zeroes P & Q
E₃ = cx² + bx + a Zeroes P & Q
P + Q = - c/b = - b/c => b² = c²
PQ = a/b = a/c => b = c
replacing c with b
E₁ = ax² + bx + b
E₂ = bx² + bx + a
E₃ = bx² + bx + a
a²/bc + b²/ca + c²/ab = 3
=> a²/b² + b²/ba + b²/ab = 3
=> (a/b)² + (b/a) + (b/a) = 3
let say a/b = y
=> y² + 2/y = 3
=> y³ - 3y + 2 = 0
=> (y - 1)²(y + 2) = 0
=> y = 1 or y = - 2
=> a/b = 1 or a/b = -2
=> a = b or a = - 2b
if using a = b
then E₁ = E ₂ = E₃ = bx² + bx + b ( hence both root common for all)
so a = - 2b
using this
E₁ = ax² + bx + b = -2bx² + bx + b = -b(2x² - x - 1) = -b(2x + 1 )(x - 1)
roots = -1/2 , 1
E₂ = E₃ = bx² + bx + a = bx² + bx -2b = b(x² + x - 2) = b(x + 2)(x - 1)
roots = - 2 & 1
1 is common roots in all
-2 is common in E₂ & E₃ but not in E₁
a) & c ) options are not free from a , b , c
option b is same as E₁ ( hence not possible)
lets check option d
x² - a(b + c)x/bc + a²/bc
using a = - 2b & c = b
= x² - (-2b)(b + b)x/b*b + (-2b)²/b*b
= x² + 4x + 4
= ( x + 2)²
hence zero = - 2
Same as common in E₂ & E₃ but not in E₁
Hence x² - a(b + c)x/bc + a²/bc is the required Quadratic equation.
( may be there is some simpler solution as well , but i could solve by this only as of now )
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1 is common zero to both E1,E2 and E3
but it is given that,
{Zeroes of E1 is not eqaul to zeroes of required eqaution
so,1 will not be the zero of the required equation..}......i).
now,
zeroes of E2:(a/b,1)
zeroes of.E3:(a/c,1)
from.this we can say
zeroes of E2 and E3 :(a/b,a/c)
not 1 because of reason in eqaution i)
now.
required eqaution=x^2-(sum of roots)x+product
=x^2-(a/b+a/c)x+a^2/bc
=x^2-a(b+c)x/bc+a^2/bc
hence option d) is the correct one
