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GED Math: Expression and Equations Drill 3, Problem 4. What are the roots of the quadratic equation?
SAT Math 8.6 Algebra and Functions 231 Views
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Description:
SAT Math 8.6 Algebra and Functions
Transcript
- 00:03
It's a bird! It's a plane! It's Shmooperman...guy!
- 00:09
Find the positive x-coordinate of the point where the graphs of the equations…
- 00:14
y = -2x squared - 6 y = -3x - 8 …intersect.
- 00:22
Okay…so this question is asking us to solve for two linear equations.
- 00:28
So, back in our old algebra days, we knew we had several methods that could help us do this…
Full Transcript
- 00:33
Solving by substitution…
- 00:35
…solving by elimination…
- 00:36
…or solving by graphing.
- 00:38
So in this case we can do any of these, but the easiest would be to just set them equal,
- 00:42
because…y equals y.
- 00:44
So we have -2x squared – 6 equals -3x – 8.
- 00:50
Now combine like terms and move everything to one side and we get:
- 00:54
2x squared – 3x – 2 = 0
- 00:59
…and that looks like a FACTORABLE quadratic equation.
- 01:03
Easiest way to solve this equation is to factor by grouping.
- 01:07
So if we split our -3x into x and -4x...
- 01:12
...we get 2x squared + x - 4x - 2 = 0.
- 01:19
This whole thing factors into x(2x + 1) - 2(2x + 1) = 0
- 01:27
Now we both have x and -2 multiplied by the quantity 2x + 1.
- 01:31
Do you know what we have to do next?
- 01:33
Factor out that 2x + 1.
- 01:36
Our final, fully factored equation is (2x + 1)(x-2) = 0.
- 01:44
Well, when x = -1/2 and 2... this equation is true.
- 01:48
Since the question asks for the positive x-coordinate, we'll take the positive 2.
- 01:54
So the answer is…2. That’s where they intersect.
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