### Video Transcript

Given that 67 over six root two plus root five is equal to π root two plus π root five, find the values of π and π.

We can observe that there are two radicals in the denominator of the fraction, whereas in the required form, these same radicals no longer appear in a denominator but are instead multiplied by constants. This suggests that in order to answer this question, we need to rationalize the denominator. We do this by multiplying both the numerator and denominator by a value that is carefully chosen because its product with the denominator gives a rational value.

As we multiply both the numerator and denominator by the same value, weβre effectively multiplying the entire fraction by one and so the result is equivalent to the original fraction. The value we choose is the conjugate of the denominator, which we obtain by changing the sign of one of the terms. Weβre going to multiply by six root two minus root five. But we could equally change the sign of the first term and multiply by negative six root two plus root five. The reason we choose this value is because the product of a number and its conjugate is always rational.

If we expand the product in the denominator, we obtain six root two squared minus six root two root five plus six root two root five minus root five squared. The two terms in the center of the expression cancel one another out, leaving six root two squared, which is 36 times two, minus root five squared, which is five. Overall, the denominator evaluates to 67, confirming that the product of this expression and its conjugate is a rational value. Before expanding the parentheses in the numerator, we can observe that there is a shared factor of 67 in the numerator and denominator. So, instead we can cancel this shared factor. Doing so gives six root two minus root five.

We now need to compare this expression with the requested form. The value of π, which is the value weβre multiplying root two by, is six. The value of π, which is the value weβre multiplying root five by, is negative one. Hence, by first rationalizing the denominator of this quotient, weβve written it in the required form and found that π is equal to six and π is equal to negative one.