Question 1

I don't know whether or not I should be posting this under this video, but I've been wondering about it for a while now. A LOOONG time ago when squares and square roots were explained, I thought that the square root of, say, 4 was just 2. But now that I am getting into more complicated things, it is never just 2 anymore it is + or - 2. Why is this? Are they different situations? Or is + or - just more correct?

Accepted Answer

+ or - is just more correct. For example, yes the square root of 4 is 2. But, it also is -2 because -2 multiplied by -2 is 4 also. Depending on the teacher or the class, you may have to write it either way. The teacher I have always wants us to write both

Question 2

What kind of weird shape does `2x^2 - y^2 = 7` make?

Accepted Answer

Wrath,
The graph would look like this: https://www.khanacademy.org/cs/hyperbola-2x-y-7/6201478981091328

Question 3

I disagree, that you can substitute back the values for x into either equation as Sal mentions at 2:30. If for example you substitute x = 2, into the 2x^2 - y^2 = 7 equation when you solve for y you get two possible values y = 1 and y = -1. This would indicate that both (2, 1) and (2, -1) are solutions, however (2, -1) is NOT a solution. By not plugging x = 2 back into the original equation y = 1/2 x, we have only insured our solutions satisfy the second equation. To see that this is the case plug (2,-1) into the first equation and you get -1 = 1/2*2, which leads to -1 = 1 which cannot be true. The reason why this isn't a problem in the second equation is that the y term is squared in that equation, which turns -1 into 1.

Accepted Answer

If you graph this, you will have a sideways-facing hyperbolic curve intersected by a straight line that passes through (-2, -1) (2,1), so those are the solutions.  It is pretty straightforward as to what the solutions are, so what's happening with the math?
My first suspicion was that there may be extraneous solutions if you substitute into the 
2x² - y² = 7

(2nd degree relationships are notorious for that, and you always need to check all constraints to see they are satisfied.)

rearranging, y² = 2x² - 7
y = ± √(2x² - 7)
For x = 2, y = ±√(2∙2² - 7)  = ±√1
For x = -2, y = ±√(2∙(-2)² - 7)  = ±√1
so y can potentially equal ±1  giving us the potential solutions of (2, 1) and (2, -1)as wll as (-2, 1) and (-2, -1)
Always with these we would be cautious of extraneous solutions,. so we have to do a lot of checking.   `Can y =-1 when x = 2?`  If we use our original equation, the matter is obscured by the squares.
So, we check with the *other* constraint, the linear equation and see if each of these really is a solution.

(2, 1) Indeed, y = 1/2 x works with (2, 1) because 1 = (1/2)2 `So (2,1) is a solution.`
(2, -1) If we try (2, -1), we get -1 =?  (1/2)2  -1 ≠ 1   `So (2, -1) is not a solution.`
(-2, 1) If we try this, we get  1 =?  (1/2)(-2)  +1 ≠ -1   `So (-2, 1) is not a solution.`
(-2, -1) Again, -1 = (1/2)(-2) because -1 = -1  `So (-2, -1) is the other solution.`

The linear equation is usually more straightforward in providing constraints and fewer extraneous answers.
So, while you can substitute into either equation, you have to beware extraneous solutions and check your answers with all the original equations, because those are constraints on your answer.  
When in doubt, it is always helpful to sketch the graphs to figure out what is going on, and having an intuitive feel for what is reasonable helps you attack the next question with more confidence.

Question 4

would angled parabolas(In other words not straight upwards or downwards or sideways) be of the form:

a * sqrt(bx) +/- c * x^2 +/- dx +/- e(not the irrational number but rather a constant)?

Accepted Answer

Consider a conventional Cartesian coordinate plane depicting a parabola of the form y=Ax²+Bx+C. Now imagine rotating every point on the plane θ degrees clockwise, resulting in a primed coordinate system depicting the exact same parabola, except this time it appears to be rotated when viewed from the reference frame of the xy coordinate plane. With respect to primed coordinates, our equation for the rotated parabola is y′=A(x′)²+B(x′)+C. In order to write this equation in terms of x and y coordinates, we will need to substitute equations of the form x′=f(x,y,θ) and y′=g(x,y,θ) into y′=A(x′)²+B(x′)+C.

Let (x′,y′) be an arbitrary point on the primed Cartesian plane that makes an angle of Φ with the x′ axis. Since the x′ axis is θ degrees below the x axis, the angle that the point (x′,y′) makes with the x-axis is Φ-θ. Hence, the x and y coordinates corresponding to (x′,y′) are x=r′cos(Φ-θ) and y=r′sin(Φ-θ), where r′=(x²+y²)^0.5. We can also write x=r′cos(Φ)cos(θ)+r′sin(Φ)sin(θ) and y=r′sin(Φ)cos(θ)-r′sin(θ)cos(Φ) because of the trig identities cos(Φ-θ)=cos(Φ)cos(θ)+sin(Φ)sin(θ) and sin(Φ-θ)=sin(Φ)cos(θ)-sin(θ)cos(Φ).

From basic trigonometry, it is clear that x′=r′cos(Φ) and y′=r′sin(Φ). This leads way to a nice simplification for the equations for x and y as follows: x=r′cos(Φ)cos(θ)+r′sin(Φ)sin(θ)=x′cos(θ)+y′sin(θ) and y=r′sin(Φ)cos(θ)-r′sin(θ)cos(Φ)=y′cos(θ)-x′sin(θ).
Solving for x and y gives us x′=xcos(θ)-ysin(θ) and y′=xsin(θ)+ycos(θ).

Finally, the rest of the derivation involves substituting x′ for xcos(θ)-ysin(θ) and y′ for xsin(θ)+ycos(θ) into y′=A(x′)²+B(x′)+C, which leads us to Acos²(θ)x²+Asin²(θ)y²-2Asin(θ)cos(θ)xy+(Bcos(θ)-sin(θ))x-(sin(θ)+cos(θ))y+C=0.

In order to convince oneself that this formula is valid, it helps to try out some cases. Suppose you have the parabola y=x². Turning it 90 degrees clockwise brings us to x=y²; turning 180 degrees brings us to y=-x²;rotating 270 degrees brings us to x=-y²; and turning 360 degrees brings us back to y=x². To confirm these observations using the formula derived, let A=1, B=0, C=0 and θ=π/2,π,3π/2, and 2π to take the different degrees of rotation into account.

Question 5

So, when a system is a linear equation and another is quadratic, that means there will be two solutions. Right?

Accepted Answer

Not always. If they never intersect (the line could be under the parabola) then there will be no solution. If they intersect at the vertex only then they will have only 1 solution and 2 solutions if they intersect twice.

Course: Algebra (all content) > Unit 9

Quadratic systems: both variables are squared

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