Answer: 9
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Step-by-step explanation:
Factor the given expression
x^2+x^3
x^2*1+x^2*x
x^2(1+x)
x^2(x+1)
We end up with something where the factor x^2 is a perfect square regardless of whatever we replace x with from the set {1,2,3,...,99,100}.
In short, we don't need to worry about the x^2 portion.
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The only thing of importance is the x+1 factor.
The x+1 must be a perfect square so that the original expression is as well.
Let's say M is some perfect square between 1 and 100. That means M belongs to the set of values
{1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
If we want M+1 to be a perfect square, then we'll need to subtract 1 from each item in that set above to counteract the "plus 1" in "M+1".
Subtract 1 from each item to end up with this modified set
{0, 3, 8, 15, 24, 35, 48, 63, 80, 99}
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Every item mentioned in that second set will make x^2+x^3 to be a perfect square.
For example, let's pick x = 15
x^2(x+1) = 15^2*(15+1)
x^2(x+1) = 15^2*16
x^2(x+1) = 15^2*4^2
x^2(x+1) = (15*4)^2
x^2(x+1) = 60^2
This confirms that x^2+x^3 is a perfect square for x = 15. I'll let you check the other values.
There are 10 values in the set {1^2,2^2,...,9^2,10^2} = {1,2,...,81,100}. There are also 10 values in the set {0,3,8, ..., 80, 99} we got earlier. The cardinality hasn't changed. However, we'll kick out 0 because it's not between 1 and 100. So we end up with 9 as our final answer.
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To summarize, there are 9 values of x that make x^2+x^3 a perfect square, where x is some whole number between 1 and 100 inclusive.