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Zeros of p(x) = x2 - 27 are
- a)± 3√3
- b)± 9√3
- c)± 7√3
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Zeros of p(x) = x2 - 27 area)± 3√3b)± 9√3c)&...
To find the zeros of the polynomial p(x) = x^2 - 27, we need to solve the equation p(x) = 0.
Setting the equation equal to zero, we have:
x^2 - 27 = 0
Adding 27 to both sides:
x^2 = 27
Taking the square root of both sides (remembering both positive and negative roots):
x = ±√27
Simplifying the square root:
x = ±√(9 * 3)
x = ±3√3
Therefore, the zeros of p(x) = x^2 - 27 are x = ±3√3.
Setting the equation equal to zero, we have:
x^2 - 27 = 0
Adding 27 to both sides:
x^2 = 27
Taking the square root of both sides (remembering both positive and negative roots):
x = ±√27
Simplifying the square root:
x = ±√(9 * 3)
x = ±3√3
Therefore, the zeros of p(x) = x^2 - 27 are x = ±3√3.
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Community Answer
Zeros of p(x) = x2 - 27 area)± 3√3b)± 9√3c)&...
x2 - 27 = 0
x2 = 27
x = √27
x = ±3√3
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Zeros of p(x) = x2 - 27 area)± 3√3b)± 9√3c)± 7√3d)None of the aboveCorrect answer is option 'A'. Can you explain this answer?
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