### Video Transcript

The diagram below shows two function machines A and B. Both machines have the same input. Work out the range of values for which the output of A is greater than the output of B.

The first thing we can do is take the diagram for A and B and turn them into expressions. For function π΄, if the input is π₯, π₯ times two is the first step. We write that as two π₯. We add one to the two π₯. We need to square everything thatβs happened so far. That means we put brackets around the two π₯ plus one and square that. The output of function π΄ is two π₯ plus one squared.

Now for function π΅, if the input is π₯, weβre using the same variable for the input of both equations because we were told that both machines have the same input. The first thing that happens in function π΅ is to square the π₯-value, π₯ squared plus one. And then we need to multiply π₯ squared plus one times four.

We put brackets around π₯ squared plus one to indicate that the four is being multiplied by the π₯ squared and the one. The output of function π΅: four times π₯ squared plus one.

We want to consider when the output of function π΄ is greater than the output of function π΅. This means we need to consider when is two π₯ plus one squared larger than four times π₯ squared plus one.

To solve this, weβll need to expand and multiply two π₯ plus one squared. We need to multiply two π₯ plus one times itself, starting with the first term, two π₯ times two π₯ plus two π₯ times one. Moving on to the second term, we have one times two π₯ and one times one. Two π₯ times two π₯ is four π₯ squared. Two π₯ times one is itself, two π₯. One times two π₯ equals two π₯, and one times one equals one. Two π₯ and two π₯ are like terms. They can be combined to equal four π₯. The expanded form of two π₯ plus one squared equals four π₯ squared plus four π₯ plus one.

We can also multiply this four across the brackets. On the right-hand side, four times π₯ squared equals four π₯ squared, and four times one equals four. We have a four π₯ squared term on both sides of the equation. If we subtract four π₯ squared from both sides of the equation, those terms cancel out.

Now we have four π₯ plus one is greater than four. And we subtract one from both sides. Four minus one is three. Four π₯ must be greater than three. And then we divide both sides by four. π₯ is greater than three-fourths. Any time π₯ is larger than three-fourths, the output of function π΄ will be greater than the output of function π΅.

Letβs do a quick test to see if this is true. Letβs solve both of these equations for π₯ equals one. One is larger than three-fourths. So we would expect the output of π΄ to be greater than the output of π΅. Two times one is two, plus one is three. Three squared equals nine. In function π΅, one squared equals one, plus one equals two. Four times two equals eight. Nine is larger than eight. The output of function π΄ was greater than the output of function π΅.