Question

Show that the roots of the equation (y – a) (y – b) + (y – b) (y – c) + (y – c) (y – a) = 0 are real and equal only when a = b = c.

Answer 1

Step-by-step explanation:

(y-a ) (y-b) + (y- b) (y-c) +(y-c) (y-a)

y^2 - ay - by + ab + y^2 - by - cy + bc + y^2 -cy -ay + ca = 0

y^2 +y^2 + y^2 - ay -ay - by - by - cy - cy + ab + bc + ca = 0

3y^2 - 2ay -2by -2cy + ab + bc + ca = 0

3y^2 - 2ay - 2ay - 2ay + a×a + a× a + a× a = 0

{ a = b = c given }

3y^2 - 6ay + a^2 + a^2 +a^2 = 0

3y^2 - 6ay + 3a^2 = 0

if roots are real and equal

d = 0

b^2 - 4ac = 0

a = 3 , b = -6a , c = 3a^2

(-6a)^2 - 4× 3 × 3a^2 = 0

36a^2 - 36a^2 = 0

0 = 0

L.H.S = R.H.S

PROVED

Answer 2

DON'T EXPECT ANSWER FROM ME DUM