Question

If a, b and c are in AP, then, the roots of the following quadratic equation are (b-c) x^2+(c-a) x+(a-b)

Answer 1

EXPLANATION.

• GIVEN

=> a, b, c are in Ap

roots of equation =

=> ( b - c)x² + ( c - a) x + ( a - b)

TO FIND ROOTS OF THE EQUATION.

according to the question,

equation = ( b - c) x² + ( c - a) x + ( a - b)

For the roots of the equation

=> D = 0

=> b² - 4ac = 0

=> ( c - a) ² - 4 ( b - c) ( a - b) = 0

=> c² + a² - 2ac - 4 [ ab - b² - ac + bc ] = 0

=> c² + a² - 2ac - 4ab + 4b² + 4ac - 4bc = 0

=> c² + a² + 2ac - 4ab + 4b² - 4bc = 0

=> c² + a² + 2ac -4b ( a + c) + 4b² = 0

=> [ ( a + c) - 2b ]² = 0

=> a + c = 2b

This is the condition of an Ap

=> a + c = 2b

Answer 2

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Since a, b, c are in AP

SO

$2b = a + c \: \: \: ( \: \: by \: \: 1 \: \: )$

$\implies \: b - c = a - b \: \: \: \: .......(2)$

The given Quadratic Equation is

$(b-c) x^2+(c-a) x+(a-b) = 0$

Comparing with the Quadratic Equation

$A {x}^{2} + Bx + C = 0$

We get

$A \: = (b - c) \:, B = (c - a) \:, C \: = (a - b)$

Now the Discriminant

$= {B}^{2} - 4AC$

$= {(c - a)}^{2} - 4(b - c)(a - b)$

$= {(c - a)}^{2} - 4(a - b)(a - b) \: \: \: \: \: (by \: \: 2 \: \: )$

$= {(c - a)}^{2} - 4 {(a - b)}^{2}$

$= {(c - a)}^{2} - { \{2(a - b)} \}^{2}$

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

$= (c + a - 2b) (c - 3a + 2b)$

$= (2b - 2b) (c - 3a + 2b) \: \: ( \: by \: 1 \: \: )$

$= 0 \times (c - 3a + 2b)$

$= 0$

Hence the roots of the given equation are real and equal

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