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