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Write solution set: a. (2x+6)(3x-2)=0 x in rational numbers b. l4x+3l=9 x in positive numbers c. x^2=81 x in real numbers
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what is this asking?
Solve each equation.
Although I'm not sure why the solution to the abs value one is limited to positive real numbers, but so be it. That just means that you throw out any negative real numbers that you find as solutions.
they are just asking for the values of x that make the equations true. There are 2 solutions for each question. (a) solve 2x + 6 = 0 and 3x - 2 = 0 rational numbers are any number that can be writtien as a fraction... (b) solve 4x + 3 = 9 and 4x + 3 = -9 (c) find the 2 numbers then when squared give 81
is b 3/2? and is c 9?
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and a 2/3 and -3?
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