Question Video: Solving a Quadratic Equation by Equating a Given Quadratic Function with a Linear Function | Nagwa Question Video: Solving a Quadratic Equation by Equating a Given Quadratic Function with a Linear Function | Nagwa

Question Video: Solving a Quadratic Equation by Equating a Given Quadratic Function with a Linear Function Mathematics

Find all values of 𝑥 where 𝑓(𝑥) = 𝑡(𝑥), given 𝑓(𝑥) = (𝑥 + 34)² and 𝑡(𝑥) = 𝑥 + 34.

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Video Transcript

Find all values of 𝑥, where 𝑓 of 𝑥 is equal to 𝑡 of 𝑥, given 𝑓 of 𝑥 equals 𝑥 plus 34 squared and 𝑡 of 𝑥 equals 𝑥 plus 34.

Before we do any calculating, let’s think about what’s happening here. 𝑓 of 𝑥 is a quadratic function and 𝑡 of 𝑥 is a linear function. We know that quadratic functions have a parabola shape. If we’re looking for the values of 𝑥, where 𝑓 of 𝑥 is equal to 𝑡 of 𝑥, that’s another way of saying where the linear function crosses the quadratic function. And there are a few cases here. There could be two solutions where the linear function crosses the quadratic function, two places. There could be one solution only one place where the linear function crosses the quadratic function, or there could be no solutions.

But instead of looking at the graph of both of these functions, we’re going to solve using algebra. And that means we want to set 𝑓 of 𝑥 equal to 𝑡 of 𝑥. We want the 𝑥-values that make the statement 𝑥 plus 34 squared equals 𝑥 plus 34. In order to solve for 𝑥, we need to get 𝑥 by itself. And so, the first thing we think will be expanding the 𝑥 plus 34 squared so that we have 𝑥 plus 34 times 𝑥 plus 34 is equal to 𝑥 plus 34. We’ll need to distribute the 𝑥 plus 34 across the 𝑥 plus 34, which gives us 𝑥 squared plus 34𝑥 plus 34𝑥 plus 1156.

Now, we can combine the terms 34𝑥 plus 34𝑥, which equals 68𝑥. To continue to solve for 𝑥, we need to get all of the 𝑥-values to one side of the equation, so we can subtract 𝑥 from both sides. But as we have a quadratic equation here, what we really want to do is set this equation equal to zero to try and solve by factoring. And that means we’ll also subtract 34 from both sides. On the right, we’ll have zero, and on the left, we’ll have 𝑥 squared plus 67𝑥 plus 1122. And now we want to try and see if we can factor this quadratic.

To do that, we need two of the factors that multiply together to equal 1122 but when added together equal positive 67. These might be larger values than you’re used to factoring. One strategy that you can use is to use a factor tree. I know that 1122 is divisible by two, which gives me another factor of 561. I think 561 is divisible by three, which gives factors of three and 187. Once we get to 187, it’s an odd value that’s not divisible by three. So, I would check seven, which doesn’t work. And then, I would check the next prime value after seven, which is 11. 11 is a factor of 187. 11 times 17 equals 187. And since 17 is a prime number, we’ve found all the prime factors of 1122.

But remember, we’re looking for two factors of 1122 that when added together equal 67. So, first, I might try to multiply two times 11 which is 22, and three times 17 which is 51. This means we can say that 22 times 51 does equal 1122, but 22 plus 51 equals 73 and not 67. So, we can try another combination of maybe two times 17 which equals 34, and three times 11 which equals 33. This means we’re saying that 34 times 33 equals 1122. And when we add 34 and 33 together, we do get 67. So, we plug in 33 and 34 and we have one factor of 𝑥 plus 33 equals zero and another factor of 𝑥 plus 34 equals zero. When we solve for 𝑥, we get 𝑥 equals negative 33 or 𝑥 equals negative 34.

Before we leave this problem completely, I wanna point out that it’s possible we could’ve found some values of 𝑥 using number sense, just by thinking about these equations. At this first step, we have 𝑥 plus 34 squared is equal to 𝑥 plus 34. We could’ve thought that negative 34 plus 34 equals zero and zero squared is equal to zero. And that means it’s possible that you could’ve recognized negative 34 as one of the factors. In addition to that, at the stage where you expanded 𝑥 plus 34 squared into 𝑥 plus 34 times 𝑥 plus 34, we could’ve divided both sides of the equation by 𝑥 plus 34. On the left, you’d be left with 𝑥 plus 34 and on the right 𝑥 plus 34 over 𝑥 plus 34 is equal to one. From there, you subtract 34 from both sides of the equation and you find the other factor of 𝑥 equals negative 33.

However, if you were using this number sense method, it could be easy to miss one of these values. If you started here, you could have thought that 𝑥 equals negative 33 was the only correct solution. Or if you recognize negative 34, you could have listed that as the only correct solution. But the method we started with will work every time and will help you find either one solution, two solutions, or no solution.

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