Here's a method to remember when, like here, you need to find a & b, given their sum (S = a+b) & product (P = ab). If you form the product of these binomials to make the quadratic equation, (x-a)(x-b) = 0, you get x² - (a+b)x + ab = x² - Sx + P = 0 Solve that by any quadratic solution method to get a & b. In the present case, that will be x² - 12x + 66 = 0 The discriminant for this is < 0, so the solutions will be two conjugate complex numbers. Let's use "complete the square." To do that here, that "66" needs to become (½·12)² = 36, so just subtract 30 from both sides: x² - 12x + 36 = -30 x - 6 = ±√(-30) x = 6 ± i√30 Finally, check that the sum & product are the given values. As usual, this is left to the student. Fred
NB. Because the original problem was completely symmetric in a & b, swapping their values in any solution is another solution. So when you got an equation for a, the same equation pertains to b, and the two solutions of that quadratic are just a & b, in either order. You could have saved yourself a lot of work, from about the middle of the video, by noticing this.
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Here's a method to remember when, like here, you need to find a & b, given their sum (S = a+b) & product (P = ab).
If you form the product of these binomials to make the quadratic equation,
(x-a)(x-b) = 0,
you get
x² - (a+b)x + ab = x² - Sx + P = 0
Solve that by any quadratic solution method to get a & b.
In the present case, that will be
x² - 12x + 66 = 0
The discriminant for this is < 0, so the solutions will be two conjugate complex numbers.
Let's use "complete the square." To do that here, that "66" needs to become (½·12)² = 36, so just subtract 30 from both sides:
x² - 12x + 36 = -30
x - 6 = ±√(-30)
x = 6 ± i√30
Finally, check that the sum & product are the given values. As usual, this is left to the student.
Fred
NB. Because the original problem was completely symmetric in a & b, swapping their values in any solution is another solution.
So when you got an equation for a, the same equation pertains to b, and the two solutions of that quadratic are just a & b, in either order.
You could have saved yourself a lot of work, from about the middle of the video, by noticing this.