Question

If the roots of the equation
(b-c)x²+(a-c)x+(a-b)=0 are equal then prove that 2b=a+c.
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What is the meaning of Xd Xd​

Answer 1

◘ Given ◘

Roots of the equation (b - c)x² + (a - c)x + (a - b) = 0 are equal.

◘ To Prove ◘

2b = a + c.

◘ Proof ◘

(b - c)x² + (a - c)x + (a - b) = 0

• As roots are equal so,  

D = b²  - 4ac = 0

$\longmapsto$ (a - c)² - 4(b - c)(a - b) = 0

$\longmapsto$ (a - c)² - 4(ab - b² - ac + bc) = 0

$\longmapsto$ (a - c)² -4ab + 4b² + 4ac - 4bc = 0

$\longmapsto$ (a - c)² - 2.(2b)(a + c) + (2b)² = 0

$\longmapsto$ [a - c + 2b]² = 0

$\longmapsto$ a - c + 2b = 0

$\longmapsto$ 2b = a + c

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► Alternate method ◄

∵ Sum of coefficients = 0,

Hence one root is 1 and the other root is (b - c) / (a - b).  

A/q,

Both the zeros are equal. So,

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

$\longmapsto$ b - c = a - b

$\longmapsto$ 2b = a + c

Hence, proved ! ✨✨

Answer 2

Given Equation -

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

To Prove -

2b = c - a.

Proof -

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

Here , as mentioned , the roots of this equation are equal .

So -

The value of the discriminant is 0 .

=> b² - 4ac = 0

b = ( a - c )

a = ( b - c )

c = ( a - b )

=> (a - c)² - 4(b - c)( a - b) = 0

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

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

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

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

=> [a + c - 2b]² = 0

=> 2b = a + c.

Hence Shown

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