Video Transcript
Express the square root of two over the square root of eight multiplied by the square root of 20 plus the square root of 36 in its simplest form.
In this question, we are asked to rewrite an expression involving the quotient and product of square roots in its simplest form. We can start by checking if any of the radicands are perfect squares. We can then see that 36 is six squared. So we can evaluate the square root of 36 by recalling that the square root of 𝑎 squared is equal to 𝑎 for any nonnegative real number 𝑎. Hence, the square root of 36 is equal to six.
Since the expression involves the quotient and product of square roots, we can begin by recalling the product and quotient rules for square roots. First, we can recall that for any nonnegative real number 𝑎 and positive real number 𝑏, we have that the square root of 𝑎 over the square root of 𝑏 is equal to the square root of 𝑎 over 𝑏. Second, we can recall that for any nonnegative real numbers 𝑎 and 𝑏, we have that the square root of 𝑎 times 𝑏 is equal to the square root of 𝑎 multiplied by the square root of 𝑏.
Since one of our factors is the quotient of two square roots, let’s start by using the quotient rule for square roots to rewrite this factor. We set 𝑎 equal to two and 𝑏 equal to eight to obtain the square root of two over eight multiplied by the square root of 20 plus six. We can then cancel the shared factor of two in the numerator and denominator to rewrite the first factor as the square root of one-quarter. We can then apply the quotient rule for square roots in reverse to take square roots of the numerator and denominator separately. This gives us the square root of one over the square root of four multiplied by the square root of 20 plus six.
We can then evaluate each of these square roots to get one-half multiplied by the square root of 20 plus six. We can then distribute the factor of one-half over the parentheses to obtain the square root of 20 over two plus three.
We might be tempted to stop here. But remember, we need to give our answer in its simplest form. This means that we need the sizes of the radicands to be as small as possible. We can do this by applying the product rule for square roots. We note that 20 is equal to four times five. So we can set 𝑎 equal to four and 𝑏 equal to five to see that the square root of 20 is equal to the square root of four times the square root of five. We can then evaluate the square root of four as two to see that the square root of 20 is equal to two times the square root of five. We can then cancel the shared factor of two to obtain three plus the square root of five, which is our final answer.
