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# Quadratic systems: a line and a parabola

A system of equations that contains one linear equation and one quadratic equations can be solved both graphically and algebraically. Each method has its pros and cons. See an example using both methods.

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• At , why does solving for x using 0 gives us the intersecting points of the two equations? I would expect that solving for x when the equation equals zero, would give us x intercepts.
• I think i might understand why:

We are not finding where y = 0, which would give the x intercepts. We are solving a system of two equations by finding the y value of one equation (i.e y = x-3) and inputting it into the second equation. Then rearranging this new equation so that it can be solved using the zero product property.

Any elaboration/correction greatly appreciated.
• At , where does he get 6 and -1 from? Does he get it from earlier in the video and I missed it or what?
• Please check out factoring by groups on this website called - Factoring quadratics: leading coefficient ≠ 1.
the short answer is; to find a common factor from the function ax^2+bx+c
- find two numbers d & e
that multiply equal a times c; -- d*e=a*c
that add up to b; -- d+e=b
In the video that are -6 & -1 as they multiply to get 3*2 and add up to -7
• Is it ok to skip the factoring and use the quadratic formula? Or is it important to learn this way?
• Both methods are valid, but learning how to factor that way is very valuable when time is limited, so it might be worthwhile to practice.
• @ we are told that as an alternative to factoring by grouping we could. How would you do that? I attempted it and did not get either answer that he did by grouping method.
𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0 ⇒ 𝑥 = (−𝑏 ± √(𝑏² − 4𝑎𝑐))∕(2𝑎)

In our case 𝑎 = 3, 𝑏 = −7, and 𝑐 = 2,
so 𝑥 = (−(−7) ± √((−7)² −4⋅3⋅2))∕(2⋅3)
= (7 ± √(49 − 24))∕6
= (7 ± √25)∕6 = (7± 5)∕6

𝑥 = (7 − 5)∕6 = 2∕6 = 1∕3
OR
𝑥 = (7 + 5)∕6 = 12∕6 = 2

So, the factorization should be (𝑥 − 1∕3)(𝑥 − 2).

But because 𝑎 = 3, the factorization is actually
3(𝑥 − 1∕3)(𝑥 − 2) = (3𝑥 − 1)(𝑥 − 2)
• At , why does he solve for y? Wouldn't it be easier to substitute y in the linear equation for the quadratic expression?
• Yes that is also another fast way to do it. There are many ways to approach a problem and still get the same solution — it's ultimately up for preference.
• Why is it x+1?!
• y-x+1=0
+x +x
-1 -1
y = x-1
(1 vote)
• At about , how does the -1 magically appear? I don't see how Sal got it.
(1 vote)
• Sal factors a -1 out of the expression. It is originally (-x + 1), and he rewrites it as -1(x - 1). You can imagine that he is 'un-distributing' a -1 to simplify the expression.
• i dont get it how do i do it
• Just do your best and eventually you will get it.
(1 vote)

• Can't you just substitute 3x^2-6x+1 for y in the linear equation?
(1 vote)
• Yes that would be quicker and more direct than what Sal did, you get to straight to 3x^2-6x+1-x+2=0, so 3x^2-7x+2=0.
(1 vote)
• Out of curiosity at , he says "two curves," even though one is a line and the other not; is this a mistake?
(1 vote)