Spud wants to find a quadratic function that has 3+4i and 3-4i as its roots. Let's try Alexia's method. According to her, a quadratic function that has two prescribed roots can have the following form.
y=(x-first root)(x-second root)
In our case, 3+4i is the first root and 3-4i is the second root. Let's substitute these roots into the equation.
y=(x-first root)(x-second root)
⇓
y=(x-( 3+4i))(x-( 3-4i))
Finally, let's simplify our equation.
Now, notice that we get a quadratic equation as a result. Since we want a quadratic function, we can replace 0 on the left hand side with y to get a function whose roots are exactly the roots of our equation. Let's do it!
0 = x^2 -6x + 25
⇓
y = x^2 -6x + 25
Extra
Why Hugo's Method Works
If we compare the results of Part A and Part B, we can see that both methods give the same answer. Let's think about why Hugo's method works. We will compare how the method works for both of the known roots.
x = 3+4i &⇔ x-3 = 4i
x = 3-4i &⇔ x-3 = -4i
The next step is to square both sides of the equation. x-3 = 4i &⇒ (x-3)^2 = (4i)^2
x-3 = -4i &⇒ (x-3)^2 = (-4i)^2
Recall that the square of a number is equal to the square of its opposite.
(4i)^2 = (-4i)^2
This means regardless of the root we start with, after this step, the resulting equation is the same.
(x-3)^2 = (4i)^2
To get to this result, we subtracted a number from both sides of the equation and squared both sides of the equation. If a number was a root of the original equation, it will still be a root after either of these two operations. For this reason, both 3+4i and 3-4i are roots of the resulting equation.
c
Both methods gave the same result.
y=x^2-6x+25
This means Spud could use either method to get the correct result. In this case, Hugo's method might be slightly quicker. However, it is important to remember that Alexia's method works better in more general scenarios — for example, when we want to find a function that has more than two known roots.
