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Find the zeroes of the polynomial 100x^2-81 |
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Answer» 100x²-81 = 0 => 100x² = 81 => x² = 81/100 => x = +/- _/81/100 => x= +/- 9/10So, the zeroes are 9/10 , -9/10 It can be solve by factorzation method |
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