Question 1

Does the square root always have to equal zero

Accepted Answer

When finding the roots, solutions, or zeros (all the same thing) to a quadratic function, yes, you set them equal to zero, and solve for the independent variable (x).

Question 2

What would the complex roots be?

Accepted Answer

Hello gatogreensleeves, 
the complex roots would be x = -5 +j*(squareroot(45)) and x = -5 -j*(squareroot(45)). 
Regards!

Question 3

At 0:52 Sal factors out a 4 to simplify the equation, which I understand. I also understand that, in the end, this equation doesn't have any real roots because we can't take the square of a negative. However, wouldn't the final answer be -45 = 4(x+5)^2 instead of -45 = (x+5)^2. The 4 that he factored out seems to have disappeared into the ether. Am I missing something?

Accepted Answer

John,
Rather that "factor out the 4", Sal divided both sides of the equation by 4.  On the left was a zero, and 0/4 = 0, so yes, the 4 did look like it evaporated into thin air.
But if he had just factored the 4, the answer would be 
-180 = 4(x+5)².  And we could still then divide both sides by 4 and the answer would be -45 = (x+5)²

0=4x² + 40x +280  Let's just factor out the 4
0=4(x² + 10x + 70) And and and subtract the 25 inside the parenthesis.
0=4(x² + 10x + 25 - 25 + 70) Now convert the perfect square
0=4((x+5)² -25+70)  subtract the 25 from 70
0=4((x+5)² + 45)  Distribute the 4
0=4(x+5)² + 4*45  and multiply 4*45
0=4(x+5)² + 180 Now subtract 180 from each side
-180 = 4(x+5)²  And now you could divide each side by 4
-45 = (x+5)²

So, as you see the 4 didn't just disappear.  It is that when one side was 0, when both sides were divided by 4 it just looked like it was factored out and disappeared.

I hope that makes it click for you.

Question 4

why can't you use x = -b + or - sqrt b squared - 4ac/2a for this?

Accepted Answer

You certainly can use that formula (which is called the quadratic formula).  However, Sal is trying to explain HOW to find the solutions of the equation so that people will understand how equations are manipulated.  Simply plugging numbers into a formula certainly works, but it doesn't help with understanding and it makes quite a boring video.

Question 5

at 4:55 what is a complex number?

Accepted Answer

Complex numbers are a new class of numbers that support roots of negative numbers.
They are written in the form a + bi where i is sqroot(-1). Note that complex numbers include the real numbers; you could think of a real number as being a complex number where b=0.
In the UK they are introduced at A-level, and used extensively in engineering degree calculations. At GCSE it is sufficient to state that there are no real roots. If you were to sketch the graph, it would never intersect the x-axis (i.e. there is no real value of x such that y=0).

Question 6

My professor said to find the vertex of the parabola y=x^2-4x+6. When you complete the square you end up getting an answer that has no solution as Sal showed above. But then why is the vertex (2,2)?

Thank You!

Accepted Answer

The vertex is different than the solution. The vertex is the maximum/minimum point of the line, and the solution(s) is where the line crosses the x-axis. The vertex can be found using this equation: x=(-b)/2a. This will get you the x value of the vertex. Plug that into your original equation to find y, and you have the coordinates of the vertex. :)

Question 7

Do we follow BODMAS in reverse for algebraic manipulation? I don't see why he couldn't have taken the square root of both sides before moving 45 to the left. He'd have had: -sqrt(45) = (x + 5).

Accepted Answer

When you perform algebraic manipulation like this, you have to remember to perform the same operation to the _entire_ side on both sides. That means taking the square root of both sides at that point, you'd have to take the square root of (x+5)^2 + 45 as a whole, not just the (x+5)^2 part. Then with the 45 "stuck" under the square root, you wouldn't be able to just subtract 45 from each side easily.

You don't _have_ to follow any particular order for which operations you perform during manipulation. Generally you need to look at what you have and see which operation will help you get closer to your goal, and remember, you have to perform the operation on the entire side as a whole for both sides.

Question 8

I see how it fits, but why did we take 1/2 of b and square it?

Accepted Answer

Great question
Sal is trying to complete the square and in order to do that he wants the right side to look something like this:
(ax + c)^2  where 'a' and 'c' can be any number
If we work backwards it might make more sense (going from (a + b)^2 to what Sal starts with).
(ax + c)^2  ---->  (ax)^2 + 2acx + c^2
We have 'a' (a = 1), and we know that b = 2ac, so to complete the square we just have to find c^2
b = 2ac ----(a = 1)---->  b = 2c  ----> b/2 = c  ---->  (b/2)^2 = c^2
I hope that helps!

Question 9

At 2:10,  Sal said we can add a billion and subtract a billion without changing the equation. I agree completely with that.
But can we add infinite and subtract infinite  without changing the equation?

Accepted Answer

No, because you can't actually add or subtract infinity.  It is not a number, it is an abstract concept.

Infinity minus infinity is not zero.  Similarly, infinity divided by infinity is not one.  Therefore, you cannot apply the same rules of inverse operations to infinity as you would to an actual number.

Question 10

When talking about the complex roots would you need a 3 dimensional graph to plot the roots? With x, y and i axes?  Or is this not possible? If not why not?

Accepted Answer

Generally complex numbers are graphed on a slightly modified coordinate plane with real numbers represented by what we usually think of as the x-axis and imaginary numbers represented by what is usually thought of as the y-axis. For example, the number 3 + 4i on the complex plane would be in the same place as the point (3, 4) on the x-y plane.

That said, I don't see why what you propose wouldn't be a reasonable representation. It'd certainly be interesting to see how the graphs look with both real and complex roots showing!

That said, I don't see why what you propose wouldn't be a reasonable representation. It'd certainly be interesting to see how the graphs look with both real and complex roots showing!

Integrated math 2

Course: Integrated math 2 > Unit 3

Solving quadratics by completing the square: no solution

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Video transcript