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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

Suraj Kumar

6 Years agoGrade

2 Answers

Arun

6 Years ago

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

Aditya Gupta

6 Years ago

once again, aruns answer is incomplete and slightly wrong.

correct method:

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 :)