Solving a 'Harvard' University entrance exam | x=? ✍️

[@brianwade4179]

At 2:01 you pulled m=5 out of the air as if by magic. This teaches the student nothing. You should have used the Rational Root Theorem on your cubic to cite the possible rational roots and then set about testing each one. You could have tested m=5 first and found it were a zero. Then you could have used polynomial division to find m^2 + 6m + 31 as the other factor. You could then have completed the square to find that your quadratic has no real roots. Instead of doing what I just said, you did a bunch of steps that amount to magic. If you really want to teach students how to solve these problems, you need to teach them how to figure them out without pulling rabbits out of hats.

[@Arctic117]

Bro, can you explain how to figure out that m=5 is a possible answer? The video doesn't include it... Explain the method please

[@brianwade4179]

@@Arctic117 Sure. Go read up on the Rational Root Theorem ("RRT"). The theorem says this. When we have a polynomial ax^n + bx^(n-1) + ... (etc.), all the way down to some constant term z, with all the coefficients being integers, if there exists a rational root for the polynomial, the root is of the form f1/f2, where f1 is a factor of the constant term (z) and f2 is a factor of the leading coefficient (a). In this specific problem the polynomial we are trying to solve is m^3 + m^2 + m^1 - 155 = 0. So, according to the theorem, if there exists a rational root for this polynomial, the root is some factor of the constant term (155) divided by some factor of the leading coefficient (1). So go find all the factors of 155. Seeing 155 ends in a 5, we easily find 155 = 5 * 31. Recognizing 31 as prime, we conclude the factors of the constant term are +-1, +-5, +-31, and +-155. The factors of the leading coefficient are +-1. So if we list out all the choices for the numerator divided by all the choices for the denominator, and then we throw away the duplicate ratios, we get that the only possible rational roots for our polynomial are +-1, +-5, +-31, or +-155. Try each of those to see which ones, if any, make the polynomial come out to zero. The only one that works is +5: 5^3 + 5^2 + 5 - 155 = 125 + 25 + 5 - 155 = 0. Thus m=5 is a root. With practice you will develop intuition about which factors to try first. Because m=5 is a root, m-5 is a factor. Now that we know m-5 is a factor, we can write our polynomial in the form (m-5)(q), where q is the quotient polynomial found by doing this computation: (m^3 + m^2 + m - 155) / (m-5). The method I use to do the computation is something known to me as "polynomial division" (PD). PD looks a lot like long division of integers. Write m-5 outside the division sign and m^3 + m^2 + m - 155 underneath the division sign. If you search in KZbin for "how to do polynomial division" you will find several videos that illustrate how it is done. In this problem, applying PD yields q = m^2 + 6m + 31 as the other factor. The last thing to check is whether m^2 + 6m + 31 factors any further. Seeing a quadratic, people tend to jump to the quadratic formula, but I find doing so to be lacking in finesse or insight. I would rather complete the square: m^2 + 6m + 9 = -31 + 9 = -22. Thus (m+3)^2 = -22, and so m+3 = +-sqrt(-22), and so m = -3 +-sqrt(-22). This does not lead to real roots, so we're done with our search. Thus m=5 is what we're looking for, and so we have 2^x = 5 which gives us x = log 5 / log 2. A good number of these "math academy", "math olympiad", or "entrance exam" videos show the content creator pulling some kind of rabbit out of his hat so as to generate a solution. You saw it happen at 2:05 when the creator somehow decided m=5. You saw it happen again at 2:55, when the creator wrote a lengthy expression for the m^3 polynomial with no explanation for how he decided what to write. The creator doing this doesn't help the viewer learn how to solve these problems. It is better to learn how to apply first principles such as RRT and PD, so you can calculate the answer without relying upon achieving magical insight.

[@AsadInternationalAcademy]

thanks

[@AsadInternationalAcademy]

Watch 110+ most important olympiad questions with solutions by just clicking on the link below kzbin.info/aero/PLybCHBiqtqWP-TrcsG21MXXxlknROi514 Don't forget to share this link with your classmates so that they also have a benefit of it. Please, like and subscribe this channel including pressing bell icon to get the notification of new video!

[@AsadInternationalAcademy]

it's easy

[@ottowotto1]

I tried to solve it by taking log base 2 of both sides to simplify it, why did this not work?

[@brianwade4179]

There's no expansion for the log of a sum.

[@ottowotto1]

@ but i didnt expand any sums every term had log base 2

[@ottowotto1]

It wasn’t log base 2 ( all three terms ) it was each individual term with log base 2

[@brianwade4179]

@@ottowotto1 Oh, you can't apply log() individually to each term. If you have LHS = RHS the next step is log(LHS) = log(RHS).

[@AsadInternationalAcademy]

You should learn this video's method to solve faster

[@kellenlockwood8206]

2^x+2^2x+2^3x=155 xln(2)+2xln(2)+3xln(2)=ln(155) ln(2)(x+2x+3x)=ln(155) 6x=ln(155)/ln(2) x=ln(155)/(6ln(2))

[@brianwade4179]

No. On LHS you can't apply ln() to each term individually. You have to apply it to the whole LHS. With LHS = RHS the next step is ln(LHS) = ln(RHS). ln(2^x + 2^2x + 2^3x) = ln(155) This approach is a dead end because there is no way to separate the terms inside the argument to ln().

[@AsadInternationalAcademy]

ok

[@AsadInternationalAcademy]

Watch 110+ most important olympiad questions with solutions by just clicking on the link below kzbin.info/aero/PLybCHBiqtqWP-TrcsG21MXXxlknROi514 Don't forget to share this link with your classmates so that they also have a benefit of it. Please, like and subscribe this channel including pressing bell icon to get the notification of new video!

[@AsadInternationalAcademy]

ok

[@AsadInternationalAcademy]

Watch 110+ most important olympiad questions with solutions by just clicking on the link below kzbin.info/aero/PLybCHBiqtqWP-TrcsG21MXXxlknROi514 Don't forget to share this link with your classmates so that they also have a benefit of it. Please, like and subscribe this channel including pressing bell icon to get the notification of new video!