The equation to solve is (1) m³ + 11m = 30 Of course you can start by bringing the constant from the right hand side over to the left hand side, but it will not be clear to many beginning students why you would then split up 30 into 8 + 22 to proceed. It is clear that the left hand side of (1) is _strictly increasing_ on the real number line, which implies that this equation can only have _one real solution_ meaning that its other two solutions will be complex. We can start looking for rational solutions using the _rational root theorem_ which guarantees that any potential rational solution of (1) must be an _integer_ which is a divisor of the constant term 30. Moreover, we can see that the left hand side is _negative_ for any negative value of m, so (1) cannot have any negative real solutions and _a fortiori_ not any negative integer solutions. Therefore, we only need to test the positive divisors of 30, which are 1, 2, 3, 5, 6, 10, 15, 30. Then, we quickly find that m = 2 is a solution since (2) 2³ + 11·2 = 30 Now, since both sides of (2) are equal, we can subtract (2) from (1) to get (m³ − 2³) + 11(m − 2) = 0 where the two terms at the left hand side have a common factor (m − 2) which we can take out to get (m − 2)(m² + 2m + 2² + 11) = 0 Of course, the complete solution of this cubic equation is now easy and can proceed as shown in your video.
The equation to solve is
(1) m³ + 11m = 30
Of course you can start by bringing the constant from the right hand side over to the left hand side, but it will not be clear to many beginning students why you would then split up 30 into 8 + 22 to proceed.
It is clear that the left hand side of (1) is _strictly increasing_ on the real number line, which implies that this equation can only have _one real solution_ meaning that its other two solutions will be complex.
We can start looking for rational solutions using the _rational root theorem_ which guarantees that any potential rational solution of (1) must be an _integer_ which is a divisor of the constant term 30. Moreover, we can see that the left hand side is _negative_ for any negative value of m, so (1) cannot have any negative real solutions and _a fortiori_ not any negative integer solutions. Therefore, we only need to test the positive divisors of 30, which are 1, 2, 3, 5, 6, 10, 15, 30. Then, we quickly find that m = 2 is a solution since
(2) 2³ + 11·2 = 30
Now, since both sides of (2) are equal, we can subtract (2) from (1) to get
(m³ − 2³) + 11(m − 2) = 0
where the two terms at the left hand side have a common factor (m − 2) which we can take out to get
(m − 2)(m² + 2m + 2² + 11) = 0
Of course, the complete solution of this cubic equation is now easy and can proceed as shown in your video.
This question frightened 300K+ examiners: m³ + 11m = 30; m =?
m³ + 11m - 30 = 0, (m³ - 2³) + (11m - 22) = 0
(m - 2)(m² + 2m + 4) + 11(m - 2) = (m - 2)(m² + 2m + 15) = 0
m - 2 = 0; m = 2 or m² + 2m + 15 = 0, (m + 1)² = - 14 = (i√14)²; m = - 1 ± i√14
Answer check:
m = 2: m³ + 11m = 2³ + 11(2) = 8 + 22 = 30; Confirmed
m = - 1 ± i√14: m² + 2m + 15 = 0, m² = - 2m - 15
m³ = - 2m² - 15m = - 2(- 2m - 15) - 15m = 30 - 11m, m³ + 11m = 30; Confirmed
Final answer:
m = 2; Two complex value roots, m = - 1 + i√14 or m = - 1 - i√14