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The roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c)=0 are equal then prove that a+bw+cw^2=0       w is imaginary cube root of unity
3 months ago

Answers : (2)

Arun
23020 Points
							
Root of quadratic equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c) = 0 are equal 
Means D = b² - 4ac = 0 for this equation, 
first we should rearrange the equation , 
(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) 
⇒x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca
⇒3x² - 2(a + b + c)x + (ab + bc + ca)
D = {2(a + b + c)}² - 4(ab + bc + ca).3 = 0
⇒4{a² + b² + c² + 2(ab + bc + ca)} -12(ab + bc + ca) = 0
⇒ a² + b² + c² - ab - bc - ca = 0
⇒2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0 
⇒(a - b)² + (b - c)² + (c - a)² = 0 
This is possible only when , a = b = c
 
3 months ago
Aditya Gupta
1717 Points
							
once again, aruns answer is incomplete and slightly wrong.
correct method:
 
Root of quadratic equation (x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c) = 0 are equal 
Means D = b² - 4ac = 0 for this equation, 
first we should rearrange the equation , 
(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) 
⇒x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca
⇒3x² - 2(a + b + c)x + (ab + bc + ca)
D = {2(a + b + c)}² - 4(ab + bc + ca).3 = 0
⇒4{a² + b² + c² + 2(ab + bc + ca)} -12(ab + bc + ca) = 0
⇒ a² + b² + c² - ab - bc - ca = 0
NOW COMES THE IMPORTANT PART:
we need to use the following identity: a² + b² + c² - ab - bc - ca = (a+bw+cw^2)(a+bw^2+cw)
so that (a+bw+cw^2)(a+bw^2+cw)= 0
which means either a+bw+cw^2= 0 or a+bw^2+cw= 0
if a+bw+cw^2= 0, we jave proved the desired result.
if a+bw^2+cw= 0, then we can take conjugate on both sides
so a+b*conj(w^2)+c*conj(w)= 0
but we know that conj(w)= w^2 and conj(w^2)= w
so a+bw+cw^2= 0
hence proved.
kindly approve yar :)
3 months ago
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