# Question Video: Finding an Unknown in a Rational Function given the Common Domain of Two Rational Functions Mathematics

If the common domain of the two functions πβ (π₯) = π₯/(π₯Β² + 64) and πβ (π₯) = β5/(π₯Β² + 11π₯ β π) is β β {β7, β4}, find the value of π.

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### Video Transcript

If the common domain of the two functions π sub one of π₯ equals π₯ over π₯ squared plus 64 and π sub two of π₯ equals negative five over π₯ squared plus 11π₯ minus π is the set of real numbers minus the set containing negative seven, negative four, find the value of π.

So weβre given the common domain of our two functions π sub one of π₯ and π sub two of π₯. So before we go any further, letβs inspect each of these functions. π sub one of π₯ and π sub two of π₯ are rational functions. That is, they are the quotient of a pair of polynomials. And we know that the domain of any rational function is the set of real numbers, not including any values of π₯ that make the denominator zero.

So what weβre going to do is actually identify the respective domains of π sub one and π sub two. To do so, weβre going to set the denominator equal to zero in each case. And those are the values of π₯ we can disregard from the domain of those functions. Then, we can use this information to identify and compare the common domain.

Letβs begin by setting π₯ squared plus 64 equal to zero. To solve for π₯, we would begin by subtracting 64 from both sides. And that gives us the equation π₯ squared equals negative 64. Our next step would be to take both the positive and negative square root of negative 64. But we know that does not give us a real value. So there are no real solutions to the equation π₯ squared plus 64 equals zero. And so the domain of π sub one of π₯ is in fact just the set of real numbers. There are no values that make the denominator equal to zero, so no values of π₯ that we have to exclude from our domain.

So what about the domain of π sub two of π₯? Once again, itβs the set of real numbers, but we need to exclude any values of π₯ that make the expression π₯ squared plus 11π₯ minus π equal to zero. Now, normally, we would try to factor the left-hand side here. But π is an unknown constant, so we canβt do that.

So instead, letβs consider the information weβve been given about the common domain. Itβs the set of real values minus the set containing negative seven and negative four. We know that there were no exclusions for π sub one of π₯. So this must mean that if π₯ is negative seven or if π₯ is equal to negative four, that makes the denominator π₯ squared plus 11π₯ minus π equal to zero.

So to find the value of π, that means that π₯ squared plus 11π₯ minus π is equal to zero when π₯ is equal to negative seven, letβs substitute π₯ equals negative seven in. When we do, we get negative seven squared plus 11 times negative seven minus π equals zero. The left-hand side simplifies to negative 28 minus π. And then adding π to both sides of the equation, we get π equals negative 28.

So we assume the value of π that makes the expression π₯ squared plus 11π₯ minus π equal to zero, where the domain also excludes π₯ equals negative seven and π₯ equals negative four, is negative 28. But letβs check by substituting π₯ equals negative four. Thatβs negative four squared plus 11 times negative four minus π equals zero, which once again simplifies to negative 28 minus π equals zero, so π equals negative 28.

Given information then about the common domain of our functions, π is negative 28.