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Course: Class 10 (Old) > Unit 4
Lesson 5: Quadratic equations word problems- Quadratic equations word problem: triangle dimensions
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Quadratic equations word problem: triangle dimensions
Sal solves a geometry problem using a quadratic equation. Created by Sal Khan and Monterey Institute for Technology and Education.
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At1:35when he says "let me make it clear... lets do it this way"
How did he know to multiply the 1/2 with the b before distributing?
Is that part of the order of operations? Are you supposed to do any multiplications that come before the parentheses first before distributing always?(28 votes)
Communative property of multiplication- You can do it in any order.(10 votes)
- How do you do factoring by grouping?(10 votes)
I'll explain how to do factoring by grouping but first be aware that.....
The quadratic equation is in the form of...........
Ax²+Bx+C=0
Where:
(note:coefficient of means the number beside)
A=the coefficient of x²
B=the coefficient of x
C=the constant term(or the term that doesn't have an x beside it)
And we want to find two coefficients of x (q and w) to replace B which when you.......
multiply them they equal = A*C =q*w
add them they equal = A+B =q*w
so from..... Ax²+Bx+C=0
you'll get.... Ax²+qx+wx+C=0
you'll group them to.. (Ax²+qx)+(wx+C)=0
you'll further simplify it with real numbers...
Okay lets have an example so that we can understand it even more.
Lets say we want to factor an equation the form of Ax²+Bx+C=0
like 2x²+13x+20=0
We know that our A=2, B=13 and C=20
So we want to find two coefficients of x (q and w) to replace B which when you.......
multiply them they equal = A*C =q*w= 2*20=40
add them they equal = A+B =13=q*w =8*5
Now we know that our q and w are..... 8 and 5 because the satisfy all the conditions we are looking for.
Lets just plug in their values to..
2x²+qx+wx+20=0
2x²+8x+5x+20=0
(2x²+8x)+(5x+20)=0 ( I grouped them apropriately so that I can factor out)
2x(x+4)+5(x+4)=0 ( I factored out a 2 and 5)
(x+4) (2x+5)=0 ( I factored out the (x+4) from 2x and 5)(23 votes)
What I am still so confused on how to solve problems using quadratic equations?(10 votes)
Can someone give me a website please? Where I can practice more examples like the one on the video.(4 votes)
So are all measurement/geometry problems using quadratics like the one in the video going to have a negative answer and a positive answer? If not, how will you know which one is the correct answer? Will one give you a different answer from the other when you put the value in the original equation?(4 votes)
With respect to Geometry, the right answer will always be positive, since it's impossible to have negative area, length, volume, etc. So if there's one positive answer and one negative answer, the correct one will be positive.
The triangle problem that Sal was doing in the video will always have one positive and one negative answer. This is because the area will always be positive making "c" (ax^2+bx+c) always negative. When "c" is negative there is only one positive solution and one negative solution.
It's also important to remember that sometimes the quadratic equation will give two positive answers or two negative answers (if you plug in x^2-5x+6 to the quadratic equation, you get an example of two positive answers, while x^2+5x+6 gives you an example of two negative answers.)
Hope that helps(7 votes)
At1:07, Sal says the area is 30 inches SQUARED equal to 1/2 the base times height, yet he wrote it as:
30 = 1/2*b(b-4)
Why didn't he write it as 30^2 like the problem said?:
30^2 = 1/2*b(b-4)(4 votes)
Area is measured in unit square. Visualize it as you want to fill a shape with tiles that measure 1 inch by 1 inch. We call these "tile" square inches or inches squared. The number of tiles that you end up using isn't squared. This is why the 30 is not squared.
See this video: https://www.khanacademy.org/math/basic-geo/basic-geo-area-and-perimeter/basic-geo-unit-squares-area/v/introduction-to-area-and-unit-squares(7 votes)
- What if you had 35n^2 + 22n + 3 = 0?
What should I do?(4 votes)- 3 only has two factors, so it is inevitable that the solution will look something like this...
(_n + 3)(_n + 1)
Now, all we have to deal with are the factors of 35 and there are four possibilities (1 & 35, 35 & 1, 5 & 7, and 7 & 5), so it shouldn't be too hard to check each one.
1.) (1n + 3)(35n + 1)
35n^2 + n + 105n + 3
35n^2 + 106n + 3 =/= 35n^2 + 22n + 3
This does not work, they are not equal
2.) (35n + 3)(n + 1)
35n^2 + 35n + 3n + 3
35n^2 + 38n + 3 =/= 35n^2 + 22n + 3
This does not work
3.) (5n + 3)(7n + 1)
35n^2 + 5n + 21n + 3
35n^2 + 26n + 3 =/= 35n^2 + 22n + 3
This does not work, so it must be the last possibility
4.) (7n + 3)(5n + 1)
35n^2 + 7n + 15n + 3
35n^2 +22n + 3 = 35n^2 + 22n + 3
They match so (7n + 3)(5n + 1) is the correct factorization.
To finish the problem...
(7n + 3)(5n +1) = 0
This means that either 7n + 3 = 0 or 5n + 1 = 0
7n + 3 = 0
7n = -3
n = -3/7
5n + 1 = 0
5n = -1
n = -1/5
n = -3/7 or -1/5(4 votes)
- 2:26why is the best way to solve a quadratic to have all the terms on one side and equal to zero?(4 votes)
By putting it equal to 0, you are then finding the roots (or zeroes or x intercepts or solutions) where the quadratic crosses the x axis. If you keep a 60 on the other side, you have to find where b^2-4b crosses the y=60 line which is okay if you are completing the square and solving by taking the square root, but not for factoring which finds the x intercepts. Even the quadratic formula requires you to have it =0 so you know a, b, and c.(4 votes)
At2:25-2:35, why is this the case? Why is the best way to solve a quadratic is to set it equal to 0?(3 votes)
The goal of solving a quadratic is to find it's roots when f(x) = 0. Roots are otherwise known as zeroes of an equation. It's not the best way; rather, it's actually the only way to solve a quadratic.
Hopefully this helps !(4 votes)
are there any quadratics that can't be factored?(2 votes)
Try x^2 + 2x + 2.
Any quadratic of the form ax^2 + bx + c can be solved using the formula ( -b +/- sqrt(D) )/2a with D= b^2-4*a*c. However, if D is less than zero it cannot be solved regularly. This requires the introduction if the imaginary number i = sqrt(-1). I think this allows us to factor all quadratics.
So short answer: yes. Long short answer: no.(5 votes)
If I was presented this problem outside the context of this chapter how would I know that quadratics would be the way to go about solving it?
Is there a tell tell sign? Because triangle doesn't scream parabola to me.(4 votes)
1:35Video transcript
The height of a triangle
is four inches less than the length of the base. The area of the triangle
is 30 inches squared. Find the height and base. Use the formula area
equals one half base times height for the
area of a triangle. OK. So let's think about
it a little bit. We have the-- let me
draw a triangle here. So this is our triangle. And let's say that the
length of this bottom side, that's the base,
let's call that b. And then this is the height. This is the height
right over here. And then the area is equal to
one half base times height. Now in this first
sentence they tell us at the height
of a triangle is four inch is less than
the length of the base. So the height is equal
to the base minus 4. That's what that first
sentence tells us. The area of the triangle
is 30 inches squared. So if we take one half
the base times the height we'll get 30 inches squared. Or we could say that
30 inches squared is equal to one half times
the base, times the height. Now instead of putting
an h in for height, we know that the height
is the same thing as 4 less than the base. So let's put that in there. 4 less than the base. And then let's see
what we get here. We get-- let me
do this in yellow. We get 30 is equal to
one half times-- let's distribute the b-- times
b, let me make it clear. So let's do it this way. Times b over 2,
times b, minus 4. I just multiplied the
one half times the b. Now let's distribute
the b over 2. So 30 is equal to b
squared over 2, be careful. b over 2 times b is
just b squared over 2. And then b over 2, times
negative 4 is negative 2b. Now just to get rid
of this fraction here let's multiply both
sides of this equation by 2. So let's multiply
that side by 2. And let's multiply
that side by 2. On the left hand
side you get 60. On the right hand side
2 times b squared over 2 is just b squared. Negative 2b times
2 is negative 4b. And now we have
a quadratic here. And the best way to
solve a quadratic-- we have a second degree
term right here-- is to get all of the terms
on one side of the equation, having them equal 0. So let's subtract 60 from
both sides of this equation. And we get 0 equal to b
squared, minus 4b, minus 60. And so what we need
to do here is just factor this thing right
now, or factor it. And then, no-- if I have
the product of some things, and that equals 0, that
means that either one or both of those things need
to be equal to 0. So we need to factor b
squared, minus 4b, minus 60. So what we want to do, we
want to find two numbers whose sum is negative 4 and
whose product is negative 60. Now, given that the
product is negative, we know there are
different signs. And this tells us that
their absolute values are going to be four apart. That one is going to be
four less than the others. So you could look at the
products of the factors of 60. 1 and 60 are too far apart. Even if you made
one of the negative, you would either get positive
59 as the sum or negative 59 as the sum. 2 and 30, still too far apart. 3 and 20, still too far apart. If you had made
one negative you'd either get negative
17 or positive 17. Then you could have 4 and
15, still too far apart. If you made one
of them negative, their sum would be either
negative 11 or positive 11. Then you have 5 and a 12,
still seems too far apart from each other. One of them is negative,
then you either have their sum being
positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want
the larger absolute magnitude number to be negative so
that their sum is negative. So if we make it
6 and negative 10 their sum will be negative
4, and their product is negative 60. So that works. So you could literally
say that this is equal to b plus 6, times
b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want
to make it very clear, is different than the b that
we're using in the equation. I just used this b
here to say, look, we're looking for two numbers
that add up to this second term right over here. It's a different b. I could have said x plus
y is equal to negative 4, and x times y is
equal to negative 60. In fact, let me do it that way
just so we don't get confused. So we could write x plus
y is equal to negative 4. And then we have x times
y is equal to negative 60. So we have b plus 6,
times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve
this right here. And then we'll go
back and show you. You could also factor
this by grouping. But just from this, we know
that either one of these is equal to zero. Either b plus 6 is equal to 0,
or b minus 10 is equal to 0. If we subtract 6 from both
sides of this equation, we get b is equal to negative 6. Or if you add 10 to both
sides of this equation, you get b is equal to 10. And those are our two solutions. You could put them
back in and verify that they satisfy
our constraints. Now the other way that
you could solve this, and we're going to
get exact same answer. Is you could just break
up this negative 4b into its constituents. So you could have broken this
up into 0 is equal to b squared. And then you could have broken
it up into plus 6b, minus 10b, minus 60. And then factor it by grouping. Group these first two terms. Group these second two terms. Just going to add them together. The first one you
could factor out a b. So you have b times b, plus 6. The second one you can
factor out a negative 10. So minus 10 times b, plus 6. All that's equal to 0. And now you can
factor out a b plus 6. So if you factor
out a b plus 6 here, you get 0 is equal to b
minus 10, times b, plus 6. We're literally just factoring
out this out of the expression. You're just left
with a b minus 10. You get the same thing that
we did in one step over here. Whatever works for you. But either way,
the solutions are either b is equal to negative
6, or b is equal to 10. And we have to be careful here. Remember, this is
a word problem. We can't just state, oh b could
be negative 6 or b could be 10. We have to think about
whether this makes sense in the context of
the actual problem. We're talking about
lengths of triangles, or lengths of the
sides of triangles. We can't have a negative length. So because of that,
the base of a triangle can't have length of negative 6. So we can cross that out. So we actually only
have one solution here. Almost made a careless mistake. Forgot that we were dealing
with the word problem. The only possible base is 10. And let's see, they say find
the height and the base. Once again, done. So the base we're saying is 10. The height is four inches less. It's b minus 4. So the height is 6. And then you can verify. The area is 6 times 10
times one half, which is 30.