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## Solving quadratics by factoring

Current time:0:00Total duration:8:21

# Quadratic equations word problem: triangle dimensions

CCSS Math: HSA.CED.A.1, HSA.REI.B.4, HSA.REI.B.4b, HSA.SSE.B.3, HSA.SSE.B.3a

## Video 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.