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SAT
Course: SAT > Unit 10
Lesson 2: Passport to advanced mathematics- Solving quadratic equations — Basic example
- Solving quadratic equations — Harder example
- Interpreting nonlinear expressions — Basic example
- Interpreting nonlinear expressions — Harder example
- Quadratic and exponential word problems — Basic example
- Quadratic and exponential word problems — Harder example
- Manipulating quadratic and exponential expressions — Basic example
- Manipulating quadratic and exponential expressions — Harder example
- Radicals and rational exponents — Basic example
- Radicals and rational exponents — Harder example
- Radical and rational equations — Basic example
- Radical and rational equations — Harder example
- Operations with rational expressions — Basic example
- Operations with rational expressions — Harder example
- Operations with polynomials — Basic example
- Operations with polynomials — Harder example
- Polynomial factors and graphs — Basic example
- Polynomial factors and graphs — Harder example
- Nonlinear equation graphs — Basic example
- Nonlinear equation graphs — Harder example
- Linear and quadratic systems — Basic example
- Linear and quadratic systems — Harder example
- Structure in expressions — Basic example
- Structure in expressions — Harder example
- Isolating quantities — Basic example
- Isolating quantities — Harder example
- Function notation — Basic example
- Function notation — Harder example
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Linear and quadratic systems — Harder example
Watch Sal work through a harder Linear and quadratic systems problem.
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In a quadratic equation of form
ax^2+bx+c=0
the product of their roots is c/a.
So you dont need to solve them to find the product of two roots(28 votes)- We could just use Vieta Theorem to find the product of the roots. It's quite more convenient.(11 votes)
Another one is that the sum of the solutions of a quadratic in that form is -b/a.(5 votes)
The SAT answer grid does not have an option for a negative answer. why is the answer here negative?(4 votes)
An answer on the SAT won't be negative if you won't be able to answer it. :)(8 votes)
- you could have solved the question without even factorizing/solving the equation as the coefficient of X^0 that is -2 is already the product of the two X co-ordinates.(6 votes)
What do you mean by thecoefficient of X^0? The constant term in the equation?(3 votes)
How do you know when to substitute and when to set the equations equal to each other? For a simple problem like this where both equations equal Y I understand, but what if they don't?(5 votes)- math is so delicious i love it(4 votes)
What are the values of:
(1) 0^0
(2) infinity^0 ?
Are they equal to 1 too?(3 votes)- 0^0 and infinity^0 both are equal to infinity.
a^0 = 1 if only a<>0(2 votes)
- how do we know when to use the quadratic formula?
like why didnt we just do 2x^2+11x-6 =0
a.b = -12
a+b = 11(2 votes)
A good rule of thumb is to spend maybe 30 seconds max on trying to think of factors. If they're not coming to you, go ahead and use the quadratic formula. The quadratic formula always works, but factoring can sometimes be faster. This means that both approaches will get you a correct answer, but how often you factor vs use quadratic formula depends on how good you are at factoring.
Here, we could factor the equation like you said, by breaking up the middle term into a +12x and a -x.(2 votes)
You said the coordinates like 2 and something else, and -1 and something else. Aren't these something elses equal to 0?(2 votes)
Is there like a catchy song or something to get this stuff to stay in my head? I am in 12th grade and am worried about getting into college! HELP WANTED!(2 votes)
you just need more practice and right direction, review to do it. When I started preparing for SAT, I was in the same condition like you.(1 vote)
Video transcript
- [Instructor] We're told
if x comma y is a solution of the system of equations above
and x is greater than zero, what is the value of xy? Pause this video and see
if you can figure this out. All right now, this is interesting. We have a system of equations
and this one looks linear. The top one looks quadratic. So what I'm going to do
is on this linear one, it looks like it's solve
for y and then I could, whatever Y is equal to, I could
substitute back here for y and then let's let me
see if I can solve for x. So just focusing on this bottom equation, we can subtract three from both sides and then that would lead us to y is equal to two times
x plus three minus three and then I could distribute out this two and so this would be equal
to 2x plus 6 then minus 3, which is, of course, if
we just combine these two, we get it's equal to 2x + 3 and so now I can take this. This is equal to y, so we could substitute this back in for y and we would get 2x + 3 is equal to 2x squared + 13x - 3. Now this looks really complex, but we're just dealing with a quadratic if we can just combine
terms onto one side, so let's get rid of the 2x and
the 3 on the left hand side, so I'm gonna subtract
that from both sides, so I'm gonna subtract
the 2x from both sides and I'm going to subtract
a three from both sides and then what that leaves me
with on the left hand side, by design, it leaves me with zero and on the right hand
side, I have 2x squared. Now 13x - 2x is going to be 11x, so plus 11x and then we have 2 -3, so that is -6. Now to solve for x here, we might wanna factor and there's techniques for doing that or we could just go straight
to the quadratic formula. If either option does
not seem familiar to you, I encourage you to review that in the algebra sections of Khan Academy, but my brain goes to quadratic formula, at least for something where the leading coefficient
is not a one over here, so the quadratic formula
of course tells us if this is of the form, ax squared, actually let me write it this way, if we're of the form ax squared + bx + c, we could see that two is equal to a, 11 is equal to b, and c is equal to -6 and so the quadratic formula, of course, is negative b plus or minus
the square root of b squared minus four ac all of that over 2a. This will give us the xes that
can satisfy this equation. Now, this would mean that this is going to be
equal to negative b, b is 11, so negative 11 plus or
minus the square root, 11 squared is 121, minus four times a, which is two times c,
which is negative six, so I could put the six here
and then make that a positive, since I'm multiplying by negative six, and then all of that over
two times a, so a is two, so two times that is
going to be equal to four and now what do we have in here? Four times two times six is equal to 48, so this is equal to 48 and so 121 + 48 is 169, so let me write it this way. Everything under the radicle is 169, which is convenient
because that is 13 squared and so this is going to simplify. It's equal to negative 11 plus or minus the square root of 169,
which is 13 over 4. Now we're going to have two roots here, but they gave us a little
bit of an extra constraint. X needs to be greater than zero. If we did a -11 - 13 over 4, that's going to give us
an x value less than zero, so the root that we're going to focus on is when we add the 13, so negative 11 + 13 is, let me
actually do the two options, - 11 + 13 is 2 over 4, so 2/4 is equal to 1/2. This is good. That is greater than zero. The other option would have been -11 - 13, which is -24, negative 24 over 4, which is equal to -6. Either of these x values would satisfy, would be a, if we think about
it, a point of intersection or the x coordinate of
the point of intersection, but it tells us x is greater than zero, so we rule this one out. So we figured out the x
value they care about, now to calculate xy, we have
to figure out the y value that they care about and the simplest place, we just have to substitute this x value back into one of these equations. I find it a lot easier to substitute back into this
one that y is equal to 2x + 3. Let me rewrite it over here maybe. Y is equal to 2x + 3, so that if x is equal to 1/2, so this is equal to two
times 1/2 plus three. Well, that's just equal to one plus three, which is equal to four. So when x is 1/2, y is equal to four, we multiply the two, so if we just take 1/2, so
I'll let me write this way, xy is going to be equal to 1/2 times four, which is going to be equal to two and we are done.