Q. 35

# Solve the followi

Given:

To find: The value of x.

Solution:

In factorization, we write the coefficient of middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the product of these two factors will be equal to the product of the coefficient of x2 and the constant term.

...... (1)

Take the LCM of the denominators.

LCM is ( x - a ) ( x - b ) ( x - c )

Now solve for (1),

⇒ a ( x - b ) ( x - c ) + b ( x - a ) ( x - c ) = 2c ( x - a ) ( x - b )

⇒( a ( x - b ) + b ( x - a ))( x - c ) = 2c ( x - a ) ( x - b )

⇒ (ax – ab + bx – ab)(x – c) = 2c (x2 – bx - ax +ab)

⇒ (ax – ab + bx – ab)(x – c) = 2cx2 – 2cbx - 2cax +2cab

⇒ ((a + b)x – 2ab)(x – c) = 2cx2 – 2c(a + b)x + 2abc

⇒ (a + b)x2 – (a + b)cx – 2abx + 2abc = 2cx2 – 2(a + b)cx +2abc

⇒ (a + b – 2c)x2 + ((a + b)c – 2ab)x = 0

⇒x[(a + b – 2c)x + ((a + b)c – 2ab)] = 0

⇒ x = o

and (a + b – 2c)x + ((a + b)c – 2ab)=0

⇒(a + b – 2c)x = - [(a + b)c – 2ab]

⇒(a + b – 2c)x = - (ac + bc – 2ab)

⇒

⇒

Hence x = 0,

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