Question Video: Finding an Unknown Coefficient in a Quadratic and a Linear Function Given That They Have an Equal Set of Zeros Mathematics • Third Year of Preparatory School

Video Transcript

The function 𝑓 of 𝑥 is equal to 𝑎 squared 𝑥 squared plus 54𝑥 plus 81 and the function 𝑔 of 𝑥 is equal to 𝑎𝑥 plus nine have the same set of zeros. Find 𝑎 and the set of zeros.

In this question, we’re given two functions 𝑓 of 𝑥 and 𝑔 of 𝑥. And we’re told that these two functions have the same set of zeros. We need to use this information to determine the value of 𝑎 and the set of zeros. We can start by recalling when we say the set of zeros of a function, we mean the set of all values of 𝑥 where the function outputs zero. Therefore, any value of 𝑥 such that 𝑔 of 𝑥 is equal to zero if we input this value into 𝑓 of 𝑥, we should also get zero. And the same should be true in reverse. Any value of 𝑥 where 𝑓 of 𝑥 is equal to zero if we input this into 𝑔 of 𝑥, we should also get zero. And since 𝑔 of 𝑥 is a linear function, it will be easier to find the zeros of 𝑔 of 𝑥. So let’s start by doing this.

We want to solve 𝑔 of 𝑥 is equal to zero, so we need to solve the equation 𝑎𝑥 plus nine is equal to zero. First, we subtract nine from both sides of the equation. This gives us that 𝑎𝑥 is equal to negative nine. And now there’s two possibilities: either the constant value of 𝑎 is equal to zero or it’s nonzero. Let’s consider each possibility separately. Let’s start with 𝑎 being equal to zero.

If we substitute this value of 𝑎 into our original function 𝑔 of 𝑥, we get that 𝑔 of 𝑥 is a constant value of nine. Since 𝑔 of 𝑥 always outputs the value of nine, there is no value of 𝑥 we can input into this function 𝑔 of 𝑥 to get zero. So the zeros of 𝑔 of 𝑥 is the empty set. And we write the zeros of the function in the following notation: 𝑧 of 𝑔. But remember, 𝑎 is also a value in our function 𝑓 of 𝑥 and 𝑓 of 𝑥 must also have the same set of zeros. If we substitute 𝑎 is equal to zero into 𝑓 of 𝑥, we get that 𝑓 of 𝑥 is equal to 54𝑥 plus 81.

This is then a linear equation. So we can find the zeros of this function by solving the equation zero is equal to 54𝑥 plus 81. We subtract 81 from both sides and then divide through by 54. 𝑥 is equal to negative 81 over 54. Therefore, the zeros of this function 𝑓 is the set containing negative 81 over 54. But this is not the same as the zeros of 𝑔 of 𝑥. We showed that 𝑔 has no zeros. Therefore, we’ve shown if 𝑎 is equal to zero, 𝑓 and 𝑔 cannot have the same set of zeros. In other words, our value of 𝑎 is not allowed to be equal to zero.

So we now only need to consider the case where 𝑎 is not equal to zero. We can now divide both sides of the equation through by 𝑎 since 𝑎 is nonzero. We get that 𝑥 is equal to negative nine over 𝑎. Therefore, 𝑥 is equal to negative nine over 𝑎 is a zero of our function 𝑔. In fact, it’s the only zero. The zeros of 𝑔 is the set containing negative nine over 𝑎.

But now we can remember we’re told in the question that 𝑔 and 𝑓 have the same set of zeros. So 𝑧 of 𝑓 is also the set containing negative nine over 𝑎. In particular, we note that 𝑧 is a quadratic function in 𝑥 and it only has one root. It has a repeated root. And we can then recall the general form for a repeated root in a quadratic equation. If ℎ of 𝑥 is a quadratic function with repeated root at 𝑟, then ℎ of 𝑥 must be equal to 𝑘 times 𝑥 minus 𝑟 squared for some value of 𝑘 not equal to zero.

Therefore, our function 𝑓 of 𝑥 must be equal to 𝑘 times 𝑥 minus negative nine over 𝑎 squared, where 𝑘 is not equal to zero. We can simplify this. Subtracting negative nine over 𝑎 is the same as adding nine over 𝑎. This gives us that 𝑓 of 𝑥 is equal to 𝑘 times 𝑥 plus nine over 𝑎 squared. Let’s now distribute the exponent over the parentheses. By using the FOIL method or otherwise, we can show that this is equal to 𝑥 squared plus 18 over 𝑎 𝑥 plus 81 over 𝑎 squared. But remember, we need to multiply all of these terms by 𝑘. Doing this, we get 𝑘𝑥 squared plus 18𝑘 over 𝑎 times 𝑥 plus 81𝑘 over 𝑎 squared.

We can now determine the value of 𝑘. To do this, we remember we’re told in the question 𝑓 of 𝑥 is 𝑎 squared 𝑥 squared plus 54𝑥 plus 81. In particular, we can compare the leading terms of these two quadratics. 𝑘𝑥 squared must be equal to 𝑎 squared 𝑥 squared. So the leading coefficients must be equal; 𝑘 must be equal to 𝑎 squared. We can now substitute 𝑘 is equal to 𝑎 squared into this expression for the function. This then gives us 𝑎 squared 𝑥 squared plus 18𝑎 squared over 𝑎 times 𝑥 plus 81𝑎 squared over 𝑎 squared. We can then simplify this expression. In the second term, 𝑎 squared divided by 𝑎 is just equal to 𝑎. And in the constant term, 𝑎 squared over 𝑎 squared is equal to one. Remember, we’ve already shown that 𝑎 is not zero. This gives us that 𝑓 of 𝑥 is equal to 𝑎 squared 𝑥 squared plus 18𝑎𝑥 plus 81.

This is almost exactly the same as the function 𝑓 of 𝑥 we’re given in the question. However, the second term of this function is 54𝑥. However, we have 18𝑎 times 𝑥. Therefore, 18 times 𝑎 must be equal to 54. And if 18𝑎 is equal to 54, 𝑎 must be equal to 54 over 18, which is three. Therefore, we’ve shown the value of 𝑎 must be equal to three. And we can substitute this into the set we found for 𝑧 of 𝑓. And since negative nine over three is just negative three, this gives us the zeros of 𝑓 is just the set containing negative three.

Therefore, we were able to show if 𝑓 of 𝑥 is equal to 𝑎 squared 𝑥 squared plus 54𝑥 plus 81 and 𝑔 of 𝑥 is 𝑎𝑥 plus nine and that these sets have the same zeros, then 𝑎 must be equal to three and the set of zeros of 𝑓 is just the set containing negative three.