Question

If the root of the quadratic equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are equal show that a=b=c.

Answer 1

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

Hence, proved , if roots of given equation are equal then, a = b = c

Answer 2

Answer:

hence proved a=b=c

Step-by-step explanation:

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