Question

Find the zeroes of x2/a+b/ac-x/b-1/c

Answer 1

Step-by-step explanation:

Given Find the zeroes of x2/a+b/ac-x/b-1/c

• We need to find the zeroes of (x^2 / a ) + (b / ac) x (-x / b) – (1/c)
•      Given (x^2 /a) + (b c / a )(- x / b) – 1/c
•           So (x^2 / a) – bc / a . x / b – 1 / c
•          So x^2 / a – c/a .x – 1/c
•     Taking LCM we get
•       So c x^2 – c^2 x – a / ac

Reference link will be

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Answer 2

Correct question:

Find zeroes (x^2/a)+(b/ac) *(-x/b) - (1/c)

Answer:

The zeroes of (x²/a) + (b/ac) × (-x/b) - (1/c) is cx² - x - a = 0

Step-by-step explanation:

(x²/a) + (b/ac) × (-x/b) - (1/c) = (x²/a) + (-x/ac) - (1/c)

Now, on taking LCM, we get,

x²/a - x/ac - 1/c = 0

∴ cx² - x - a = 0

On solving the above equation, we get the zeroes of the polynomial.

The zeroes of the polynomial are the value of the variables that makes the equation zero.