Question

find the zeros of quadratic polynomial 49x²-81

Answer 1

⭐Hey here is your answer ⭐

49x^2 - 81 = 0

(7x)^2 - (9)^2

(7x + 9)(7x - 9)

either ,

x = -9/7 or x = 9/7



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Answer 2

Hey here is your answer
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49x²-81 = 0
${(7x)}^{2} - {(9)}^{2} = 0 \\ (7x - 9)(7x + 9) = 0 \\ 7x - 9 = 0 \: \: and \: \: 7x + 9 = 0 \\ x = \frac{9}{7} \: \: and \: \: x = \frac{ - 9}{7}$
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