rationalize this numerator so that [(√2(x+delta x)-3) - (√2x-3)]/delta x = 2/[(√2(x+delta x)-3) + √2x-3)]
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Hello, yess! Is there a typo on the problem? . This is not true . . .
Originally Posted by Soroban Hello, yess! Is there a typo on the problem? This is not true . . . ohh oops the square root should be over the whole expression so square root of 2(x+delta x) -3 minus the square root of 2x-3 all over delta x equals 2 over the square root 2(x+delta x)-3 + square root of 2x-3.
if you mean you "rationalize the numerator" by multiplying by
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