Question

5. If the roots of the equations (a-b) x^ 2 + (b - c) * x + (a - b) = 0 are real and equal, then which of the following is true? (a) 2b = a + c (c) 2c = a + b (b) 2a = b + c (d) 2b = a - c​

Answer 1

Answer :

option (b) : 2a = b + c

Step-by-step explanation :

➤ Quadratic Polynomials :

    ✯ It is a polynomial of degree 2

    ✯ General form :

              ax² + bx + c  = 0

                $\boxed{\bf x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

             

    ✯ Determinant, D = b² - 4ac

    ✯ Based on the value of Determinant, we can define the nature of roots.

            D > 0 ; real and unequal roots

            D = 0 ; real and equal roots

            D < 0 ; no real roots i.e., imaginary

    ✯ Relationship between zeroes and coefficients :

              ✩ Sum of zeroes = -b/a

              ✩ Product of zeroes = c/a

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Given

• Polynomial, (a-b)x² + (b-c)x + (a-b) = 0

         It is of the form ax² + bx + c = 0

          a = (a-b) , b = (b-c) , c = (a-b)

• Roots are real and equal.

Since roots are real and equal,

Determinant, D = 0

b² - 4ac = 0

(b - c)² - 4(a - b)(a - b) = 0

(b - c)² - 4(a - b)² = 0

(b - c)² - [2(a - b)]² = 0

[b - c + 2(a - b)] [b - c - 2(a - b) ] = 0

(b - c + 2a - 2b) (b - c - 2a + 2b) = 0

(2a - b - c) (3b - 2a - c) = 0

(2a - b - c) = 0/(3b - 2a - c)

2a - b - c = 0

⇒ 2a = b + c