Question

if the roots of the equation (b-c)x^2+(c-a)+(a-b)=0 are equal then a,b,c are in
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Answer 1

Answer:

a , b , c are in AP .

Solution:

Here,

The given quadratic equation is ;

(b - c)x² + (c - a)x + (a - b) = 0

Thus,

=> (b - c)x² + (c - a)x + (a - b) = 0

=> (b - c)x² - (-c + a)x + (a - b) = 0

=> (b - c)x² - (a - c)x + (a - b) = 0

=> (b - c)x² - (a - b + b - c)x + (a - b) = 0

=> (b - c)x² - [ (a - b)x + (b - c) ]x + (a - b) = 0

=> (b - c)x² - (a - b)x - (b - c)x + (a - b) = 0

=> x[ (b - c)x - (a - b) ] - [ (b - c)x - (a - b) ] =0

=> [ (b - c)x - (a - b) ]•(x - 1) = 0

=> x = (a - b)/(b - c) , x = 1

But,

According to the question , both the roots of the given quadratic equation are equal .

Thus,

=> (a - b)/(b - c) = 1

=> a - b = b - c

=> a + c = b + b

=> 2b = a + c { condition for AP }

Hence,

a , b , c are in AP.

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Another method :

Here,

The given quadratic equation is ;

(b - c)x² + (c - a)x + (a - b) = 0

On comparing with the general form of the quadratic equation Ax² + Bx + C , we have ;

A = (b - c)

B = (c - a)

C = (a - b)

Also,

We know that ,

For equal roots , the discriminant of the quadratic equation must be equal to zero .

Thus,

=> Discriminant , D = 0

=> B² - 4A•C = 0

=> (c - a)² - 4(b - c)(a - b) = 0

=> (c² + a² - 2ac) - 4(ba - b² - ca + cb) = 0

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

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

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

=> (a + c)² - 4b(a + c) + (2b)² = 0

=> (a + c)² - 2×(a + c)×b + (2b)² = 0

=> [ (a + b) - 2b ]² = 0

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

=> a + b = 2b

=> 2b = a + b { condition for AP }

Hence,

a , b , c are in AP .