working on solving equations...x2-32=0 can u help , Please?
Hi, Tracy! We want to solve for what "x" must equal to make the given equation true. In order to do that, we want to "undo" all of the things that are done to that "x" value, keeping in mind that if we do something to one side of an equation (add, subtract, multiply, take the square root, etc.), we have to do it to both sides, to keep the equation true. In this case, we have:
x^{2} - 32 = 0
First, let's "undo" the subtraction of 32 by adding 32 to both sides.
x^{2} - 32 = 0
+ 32 +32
x^{2} = 32
Now, we can "undo" the squaring of x by taking the square root of both sides:
√(x^{2}) = √(32)
We must be careful here, because anytime we take the square root of a number, the result can be either the positive OR negative version of that number...
x = +/- √(32)
So, x could be either the positive or negative square root of 32. Either of these numbers will make our original equation true. We can also further simplify the equation by factoring square numbers out of 32:
x = +/- (√(16)*√(2))
x = +/- 4√(2)
x^{2} - 32 = 0
First, let's "undo" the subtraction of 32 by adding 32 to both sides.
x^{2} - 32 = 0
+ 32 +32
x^{2} = 32
Now, we can "undo" the squaring of x by taking the square root of both sides:
√(x^{2}) = √(32)
We must be careful here, because anytime we take the square root of a number, the result can be either the positive OR negative version of that number...
x = +/- √(32)
So, x could be either the positive or negative square root of 32. Either of these numbers will make our original equation true. We can also further simplify the equation by factoring square numbers out of 32:
x = +/- (√(16)*√(2))
x = +/- 4√(2)