if the root of the equation (b-c)x² + (c-a)x + (a - b) = 0 are equal prove that 2b = a + c
Answers
EXPLANATION.
Roots of the equation.
⇒ (b - c)x² + (c - a)x + (a - b) = 0.
As we know that,
⇒ D = Discriminant Or b² - 4ac.
For equal roots : D = 0.
⇒ (c - a)² - 4(b - c)(a - b) = 0.
As we know that,
Formula of :
⇒ (x - y)² = x² + y² - 2xy.
Using this formula in equation, we get.
⇒ (c² + a² - 2ac) - 4(ab - b² - ac + bc) = 0.
⇒ c² + a² - 2ac - 4ab + b² + 4ac - 4bc = 0.
⇒ a² + b² + c² - 4ab - 4bc - 2ac = 0.
⇒ (c + a - 2b)² = 0.
⇒ c + a - 2b = 0.
⇒ 2b = a + c.
MORE INFORMATION.
Nature of the roots of the quadratic expression.
(1) = Real and unequal, if b² - 4ac > 0.
(2) = Rational and different, if b² - 4ac is a perfect square.
(3) = Real and equal, if b² - 4ac = 0.
(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.
Answer:
.Roots of the equation.
→ (b - c)x² + (c - a)x + (a - b) = 0.
→ D = Discriminant Or b² - 4ac.
D = Discriminant Or b² - 4ac.For equal roots : D = 0.
→ (c - a)² - 4(b - c)(a - b) = 0.
Formula of :
→(x - y)² = x² + y² - 2xy.
(x - y)² = x² + y² - 2xy.Using this formula in equation, we get.
→ (c² + a² - 2ac) - 4(ab - b² - ac + bc) = 0.
→ c² + a² - 2ac - 4ab + b² + 4ac - 4bc = 0.
→ a² + b² + c² - 4ab - 4bc - 2ac = 0.
→ (c + a - 2b)² = 0.
→ c + a - 2b = 0.
→ 2b = a + c.
2b = a + c. hope it helps you mate