Harvard University Entrance Exam Question | Can you solve ?

Added: 2026-06-07

[00:00:01]Hello, welcome back once again.

[00:00:03]Today, we're going to solve this interesting problem from Harvard University.

[00:00:08]We're given an equation in exponential form with a quadratic base and a quadratic exponent.

[00:00:16]In this video, we're going to find all possible real values of x that will satisfy this equation. So, let's get started.

[00:00:23]Now, we'll be considering the following cases. Let's start with our first case.

[00:00:28]So, in our first case here, this is where we're going to observe the following, that a raised to the power of 0 is equal to 1, provided that a is not equal to 0.

[00:00:51]So, this right over here implies that if this exponent here is equal to 0, therefore, this raised to the power of 0 must be 1. Hence, the exponent x squared - 13 x + 42 is going to be equal to 0.

[00:01:15]So, from here, we look for two numbers that the product is 42 and their sum is 13.

[00:01:20]So, the numbers are 7 and 6, right? So, here we write -13x as -7x - 6x.

[00:01:32]And then, + 42.

[00:01:35]This is equal to 0.

[00:01:36]So, here we factorize. So, we get x into bracket x - 7 from the first two terms, then - 6 here into bracket x - 7. This is equal to 0.

[00:01:48]So, this left-hand side is factored into x - 6, then multiplied by x - 7 is equal to 0.

[00:01:56]And we can see from here that if x - 6 is equal to zero, we get x is equal to six.

[00:02:03]If x - 7 is equal to zero, we get x is equal to seven. So, here we have two solutions for x, x = 6 and x = 7.

[00:02:11]Awesome.

[00:02:13]Now, let's see our second case.

[00:02:18]So, in our second case here, so this is where we're going to consider that 1 to the power of k is equal to one for all k, right?

[00:02:28]So, it implies that if the base is equal to one, then one to the power of the exponent must be one. Hence, the base, which is x squared - 7x + 11 is going to be equal to one.

[00:02:45]Subtract one from both sides to get x squared - 7x + 11 - 1 is 10 is equal to zero.

[00:02:54]Again, we're going to look for two numbers that are product to 10 and their sum is seven. So, we have two and five for such numbers. So, x squared - 2x - 5x then + 10 is equal to zero.

[00:03:09]So, from here we factor out x here to get x - 2 here, -5 to get x - 2. This is equal to zero.

[00:03:19]So, from here the left-hand side is factored into x - 2 then x - 5 is equal to zero. And from here, we have another two solutions, x = 2 and x = 5. Awesome.

[00:03:36]Then, for the final case here, which is our third case. So, in our third case here, this is where we're going to consider that -1 to the even power must be equal to one here for all k. Or here, for k is an integer, right? I think so. That is the best. So, here we're going to let the base x squared minus 7x plus 11 equals -1. So, add one to both sides to get x squared minus 7x plus 12 is equal to zero. Look for two numbers that product is 12. Their sum is seven. And from here, we see that x is going to be equal to three and x is going to be equal to four.

[00:04:26]Because we can write -7x as -3x -4x. And when you factor out, you will still arrive at getting x equals three and x equals four.

[00:04:36]So, therefore, in our conclusion here, we have the following solutions for x.

[00:04:42]So, we have x is equal to the following values: two, three, four, five, six, and seven.

[00:04:56]So, here we have six good solutions that satisfy the given equation. Is that not awesome? That is indeed awesome. Thank you for watching. If you enjoyed the video, please kindly subscribe to this channel. Also, like, comment, and share.