Can You Find x? A Nice Algebra Germany Olympiad Challenge

[00:00:00]Here is today's problem. We want to solve the equation xqub - x^2 = 100.

[00:00:06]Here x cub means x * x * x multiplied out in full and x² means x * x. The same number taken twice. On the right side sits the plane number 100, our target.

[00:00:18]We are looking for every value of x that makes the two sides equal. Let us start by putting the equation into a friendlier shape. The first step is to bring every term over to one side. We subtract 100 from both sides of the equation. Subtracting the same amount from both sides keeps the balance. The equation keeps the same solutions after this move. On the right, 100 - 100 leaves us with zero. That gives x cub - x^2 - 100= 0. Now the entire expression on the left is set equal to zero. Having 0 alone on one side sets the stage for factoring. On the left we now have a cubic a polomial of degree 3. Our plan is to break this cubic apart into simpler factors. Factoring trades one big equation for a few small ones. To get the factoring started, we first search for a single root. A root is a value of x that makes the whole polomial equal zero. For this search, we lean on the rational root theorem. Our cubic has whole number coefficients. So the theorem applies. The theorem describes every rational number that could be a root. Imagine a possible root written as a fraction in lowest terms. The theorem says the top must divide the constant term and the bottom of the fraction must divide the leading coefficient. In our polomial the constant term is minus 100.

[00:01:36]Its divisor are exactly the same as the divisor of 100. The leading coefficient the number in front of x cubed is 1. So the bottom of any candidate fraction has to divide one. That forces the bottom to be one. So every candidate is a whole number and that whole number has to divide 100 with no remainder. So our full candidate list is simply the divisor of 100. Divisers are the numbers that fit into 100 with nothing left over. The positive ones are 1 2 4 5 10 20 25 50 and 100. Each of these also brings along a negative partner. A number like three never appears since three does not divide 100. These are the only rational numbers that could possibly be roots. We can settle all of the negative candidates in one stroke.

[00:02:23]Suppose for a moment that x were some negative number. Then x cubed is negative. Since three negative factors give a negative meanwhile x^2 is positive. So - x^2 is negative. Negative plus negative stays negative withever the sizes are. But a solution must make x cub - x^2 = 100. A negative total can never equal the positive number 100. So every negative candidate is ruled out in a single step. Only the nine positive divisor are left for us to test. To test a candidate, we plug it into x cub minus x^2. If the output is exactly 100, that candidate solves the equation. We begin with the smallest candidate on the list, x= 1. 1 cubed is 1 and 1 2 is 1 as well.

[00:03:10]So the expression gives 1 minus 1, which comes out to zero. 0 is a long way from 100. So we cross off one and move on.

[00:03:18]The next candidate up the list is x = 2.

[00:03:21]2 cubed is 8 since 2 * 2 * 2 = 8. 2 ^2 is 4 and 8 - 4 leaves us with only 4.

[00:03:28]That is still nowhere near 100. So, we cross off the two as well. We continue upward with the next candidate. X= 4. 4 cubed is 64 because 4 * 4 is 16 and 16 * 4 is 64. 4 squared is 16. So, we compute 64 - 16 and get 48. 48 sits closer to the goal, but it still falls short of 100. So, we move one step further and test x= 5. 5 cubed is 5 * 5 * 5. So, 25 * 5, which is 125. 5^ 2ar is 25. So, the expression reads 125 - 25.

[00:04:07]125 - 25 comes out to exactly 100 right on the mark. So, x= 5 satisfies the original equation perfectly. We have found our first solution. X= 5. The value five turns the polomial into zero exactly as a root should. A polomial of degree 3 can have as many as three roots. So the equation might still be holding on to other solutions. The candidates we never reach were 10, 20, 25, 50, and 100. Instead of plugging in each one, we can factor the cubic. The factored form will judge all of them in a single move. It will also expose any roots that are not whole numbers. This is where the factor theorem comes into play for us. Since five is a root, x - 5 must divide the cubic evenly. So, we divide xqub - x^2 - 100 by x - 5. We will use long division just like long division with numbers. Our cubic is missing a plain x term in the middle.

[00:05:04]So, we rewrite it with a placeholder term plus 0x. Nothing about the polomial changes since 0x is just zero. The dividend becomes x cub - x^2 + 0x - 100.

[00:05:17]The placeholder keeps every column of the division lined up. Long division here works one power of x at a time.

[00:05:23]First we ask how many times x fits into x cub. x cub / x is x^2. Our first quotient term. Now we multiply x^2 by the divisor x - 5. That product comes out as xub - 5 x^2. We subtract this product from the matching terms above it. xqub - x cub cancels away completely to nothing. Then we handle - x^2 - -5x^2.

[00:05:48]Subtracting - 5x^2 is the same as adding 5x^2. So - x^2 + 5 x^2 leaves positive for x^2. Next, we bring down the placeholder term + 0x. Our new working line reads for x^2 + 0x. We divide for x^2 by x and obtain for x for the quotient. So far the quotient we are building reads x^2 + 4x. Multiplying for x by x - 5 gives 4x^2 - 20x. We subtract and the 4x^2 terms cancel each other.

[00:06:19]Then 0x - -20x becomes positive 20x. We bring down the final term of the dividend - 100. The bottom line of our division now reads 20x - 100. Dividing 20x gives 20. The last term of the quotient, we multiply 20 by x - 5 and get exactly 20 x - 100. Subtracting one line from the other wipes out every term. The remainder of the whole division is exactly zero. A remainder of 0 confirms that x - 5 divides cleanly and the quotient we have collected is x^2 + 4 by + 20. So the cubic splits into x - 5 * x^2 + for x + 20. Let us expand this product once and see that it matches. X * the bracket gives X cub + 4 X^2 + 20X - 5 * the bracket gives - 5 X^2 - 20 by - 100. Now we add these two rows of terms together column by column for x^2 - 5x^2 combines into - x^2. 20x - 20x cancels while the - 100 remains.

[00:07:24]So the expansion returns xqub - x^2 - 100 that is precisely our polomial. So the factorization is correct. Our equation now says this product of two factors equals zero. A product is zero exactly when at least one of its factors is zero. This principle is what makes the factored form so useful. So we take each factor in turn and set it equal to zero. The first factor gives the equation x - 5= 0. Adding 5 to both sides, we find x= 5 once again. That matches the solution we discovered earlier by testing. The second factor gives x^2 + 4 for x + 20= 0. This is a quadratic and we solve it by completing the square. Its x^2 term has coefficient one. So no dividing is needed.

[00:08:10]Completing the square rewrites it so x appears only once. We focus on the first two terms x^2 + 4 x. We take the coefficient four, cut it in half and square the result. Half of four is 2 and squaring that two produces four. So x^2 + 4 x= x + 2. all^ 2 - 4. We substitute this back into the quadratic equation.

[00:08:32]It becomes x + 2 all^ 2 -4 + 20 = 0. The loose numbers -4 + 20 combine into + 16.

[00:08:41]So the equation reads x + 2 all^ 2 + 16 = 0. Moving the 16 across, x + 2 all^2 = -16. But the square of a real number is never negative. So this factor contributes no real solutions. That leaves x= 5 as the only solution of the equation. Every root of the cubic had to come from one of those two factors. So the solution set of our equation is the single value 5. In other words, x= 5 and no other number gets the job done. To finish, we verify the solution inside the original equation. We substitute x= 5 into x - x^2. That means computing 5 cubed and 5 squared separately first. 5 cubed means 5 * 5 * 5 taken one product at a time. 5 * 5 is 25 and then 25 * 5 is 125. Next 5^ 2 is simply 5 * 5 which is 25. So the left side of the equation becomes 125 - 25.

[00:09:45]125 - 25 = 100. The exact number on the right side. So substituting five for x turns the equation into 100 = 100. Both sides agree and the solution is confirmed. We can also test it inside the factored version of the cubic. With x= 5, the factor x - 5 becomes 5 - 5, so 0. The other bracket becomes 25 + 20 + 20, which is 65. And 0 * 65 is 0 matching the rearranged equation. So the check succeeds in the factored form as well. Everything fits together and the problem is solved. So the final answer to today's problem is x= 5. Thank you so much for watching all the way to the end. If this solution was helpful, a like on the video is appreciated and subscribing to the channel means catching the next problem