Question #f992a

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Question #f992a

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Answer 1 · 2016-07-04T00:00:09

The equation has infinitely many solutions. See below.

Explanation:

The important thing to realize in order to solve this problem is that if square roots are multiplied, you can group them together. In math terms:
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$ $sqrt(a)*sqrt(b)=sqrt(ab)$

So, for example, $\sqrt{32} \cdot \sqrt{2} = \sqrt{32 \cdot 2} = \sqrt{64} = 8$ $sqrt(32)*sqrt(2)=sqrt(32*2)=sqrt(64)=8$ .

In this equation, we have $\sqrt{n^{2} - 1} \cdot \sqrt{n^{2} + 1}$ $sqrt(n^2-1)*sqrt(n^2+1)$ . Using the rule described above, we can combine them into one square root:
$\sqrt{n^{2} - 1} \cdot \sqrt{n^{2} + 1} = \sqrt{(n^{2} - 1) (n^{2} + 1)}$ $sqrt(n^2-1)*sqrt(n^2+1)=sqrt((n^2-1)(n^2+1))$

The expression $(n^{2} - 1) (n^{2} + 1)$ $(n^2-1)(n^2+1)$ may not look like anything we know at first, but recall the difference of squares property:
$(a - b) (a + b) = a^{2} - b^{2}$ $(a-b)(a+b)=a^2-b^2$

$(n^{2} - 1) (n^{2} + 1)$ $(n^2-1)(n^2+1)$ is actually a difference of squares, with $a = n^{2}$ $a=n^2$ and $b = 1$ $b=1$ . Since this simplifies into $a^{2} - b^{2}$ $a^2-b^2$ , we can say:
$(n^{2} - 1) (n^{2} + 1) = {(n^{2})}^{2} - {(1)}^{2} = n^{4} - 1$ $(n^2-1)(n^2+1)=(n^2)^2-(1)^2=n^4-1$

We've just simplified $\sqrt{n^{2} - 1} \cdot \sqrt{n^{2} + 1}$ $sqrt(n^2-1)*sqrt(n^2+1)$ into $\sqrt{n^{4} - 1}$ $sqrt(n^4-1)$ . Our problem now looks like:
$\sqrt{n^{4} - 1} = \sqrt{n^{4} - 1}$ $sqrt(n^4-1)=sqrt(n^4-1)$

Hm...both sides are the same! What does that mean? It means the equation has infinitely many solutions! Any value of $n$ $n$ will satisfy this equation.

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Question #f992a

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