What are the correct values of #x# in the equation #4x^2 = y# when #y = 144#?

1 Answer
Júlio B.
Dec 28, 2015

x' = +6; x" = -6

Explanation:

First, we pass the "4" that multiplies x to divide 144: #x² = 144 / 4 = 36#
Then, we have to pass the square of x to the other side, with its value inverted: #x² = 36# >> #x = 36 ^(1/2) = sqrt(36) = +- 6 # . So, the first value of X is +6, and the second is -6

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