prove that both the roots of the equation (x-a)x-b)+(x-b)(x-c)+(x-c)(x-a)=1 are real but they are equal only when a=b=c
Answers
EXPLANATION.
Roots of the equation.
⇒ (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a).
Are real but they are equal only when : a = b = c.
As we know that,
We can write equation as,
⇒ [x² - bx - ax + ab] + [x² - cx - bx + bc] + [x² - ax - cx + ac] = 0.
⇒ x² - bx - ax + ab + x² - cx - bx + bc + x² - ax - cx + ac = 0.
⇒ x² + x² + x² - ax - ax - bx - bx - cx - cx + ab + bc + ac = 0.
⇒ 3x² - 2ax - 2bx - 2cx + ab + bc + ac = 0.
⇒ 3x² - 2(a + b + c)x + (ab + bc + ac) = 0.
As we know that,
D = discriminant Or b² - 4ac.
Using this formula in the equation, we get.
⇒ [-2(a + b + c)² - 4(3)(ab + bc + ac)] = 0.
⇒ [4(a² + b² + c² + 2ab + 2bc + 2ac) - 12[ab + bc + ca] = 0.
⇒ 4[a² + b² + c² + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca] = 0.
⇒ 4[a² + b² + c² - ab - bc - ca] = 0.
⇒ 2[2a² + 2b² + 2c² - 2ab - 2bc - 2ca] = 0.
⇒ 2[(a - b)² + (b - c)² + (c - a)²] = 0.
Roots of equation are real and equal : D = 0.
⇒ a = b = c.
Hence Proved.
MORE INFORMATION.
Nature of the roots of quadratic expression.
(1) Roots are real and unequal, if b² - 4ac > 0.
(2) Roots are rational and different, if b² - 4ac is a perfect square.
(3) Roots are real and equal, if b² - 4ac = 0.
(4) If D < 0 Roots are imaginary and unequal Or complex conjugate.