### Video Transcript

A parallelogram and a rectangle are
shown with measurements in centimetres. The area of the parallelogram is
greater than the area of the rectangle. Find all possible values of 𝑥.

First of all, to solve this
problem, what we need to do is actually see how we’re gonna find out the area of our
parallelogram and our rectangle. Well, the area of a parallelogram
is equal to base times height. And in this case, it’s actually
base times perpendicular height. And, actually, the area of our
rectangle is the same because it’s actually equal to the base times the height.

Okay, great! So now let’s use this to actually
set up an inequality. Okay, so we have the inequality
three 𝑥 plus six multiplied by 𝑥 minus one, because that’s our base multiplied by
perpendicular height, is greater than 𝑥 multiplied by 𝑥 plus four. Again, this is cause this is the
base multiplied by the height of our rectangle. And we know that it’s gonna be
greater than, cause if we take a look at the question, we can see that the area of
the parallelogram is greater than the area of the rectangle.

Okay, so now what we need to do is
actually expand our brackets. So first of all, we actually have
three 𝑥 multiplied by 𝑥, which gives us three 𝑥 squared, and then three 𝑥
multiplied by negative one, which gives us negative three 𝑥, and then six
multiplied by 𝑥, which gives us plus six 𝑥, and then finally positive six
multiplied by negative one, which gives us negative six.

Okay, great! That’s the left-hand side. On the right-hand side, we’re gonna
have 𝑥 squared, cause we’ve got 𝑥 multiplied by 𝑥. And then we have 𝑥 multiplied by
four, which gives us positive four 𝑥.

So now what I’m gonna do is
actually simplify both sides of the inequality. So that gives me three 𝑥 squared
plus three 𝑥 minus six is greater than 𝑥 squared plus four 𝑥. And we got that because we had
negative three 𝑥 plus six 𝑥, which gives us positive three 𝑥. And then what we actually do is
subtract 𝑥 squared from each side. So that gives us two 𝑥 squared
plus three 𝑥 minus six is greater than four 𝑥. So then we subtract four 𝑥 from
each side. And we get two 𝑥 squared minus 𝑥
minus six is greater than zero.

Okay, great! So now what we want to do is
actually to solve this inequality. So in order to actually solve this
inequality, what we’re gonna do first is actually find the roots of our quadratic
equation. And that quadratic equation is that
two 𝑥 squared minus 𝑥 minus six is equal to zero.

And to enable us to do that, what
we’re gonna do is actually gonna factorise. And I’m gonna show you a method how
to factorise a quadratic where the coefficient of 𝑥 squared is actually greater
than one. You can do it using trial and
error. But I can show this method because
this will work every time if you have one of these problems.

So if we think about the
coefficient of 𝑥 squared being 𝑎, the coefficient of 𝑥 is 𝑏 and then 𝑐 is the
number at the end. Then what we’re gonna do next,
well, first of all, we’re gonna multiply 𝑎 and 𝑐 together. Well, 𝑎 multiplied by 𝑐 is gonna
be two multiplied by negative six, which gives us negative 12.

And now what we want to do is we
actually want to find two numbers whose product is negative 12, so 𝑎𝑐, and whose
sum is actually 𝑏, so in this case negative one. So what we can do at this stage is
actually list different pairs of factors that would actually give us negative 12 as
the products and negative one as the sum. And if we did that, we’d arrive at
these two factors. And those two factors are negative
four and positive three. That’s because if we have negative
four multiplied by three, what we’re gonna get is negative 12. And if we do negative four plus
three, what we’re gonna have is negative one. So these are gonna be the two
factors that we’re gonna use.

So now the next step is to actually
split up our coefficient of 𝑥. So we’ve got negative four and plus
three and now rewrite it. So what we’ve done here is we’ve
got two 𝑥 squared minus four 𝑥 plus three 𝑥 minus six is equal to zero.

Now the next step is to actually
factorise each pair of terms. So we’re gonna start with the first
two terms. So we’re gonna factorise two 𝑥
squared minus four 𝑥. And if we factorise that, we’re
gonna get two 𝑥 outside the bracket because there’s a two 𝑥 in both two 𝑥 squared
and negative four 𝑥. And then inside the bracket, we’re
gonna have 𝑥, because two 𝑥 multiplied by 𝑥 is two 𝑥 squared. And we’re gonna have negative two,
cause two 𝑥 multiplied by negative two gives us negative four 𝑥.

Okay, great! So we factorised the first two
terms. So now what we’re gonna do is
actually factorise the second two terms. And if we factorise the second two
terms, we’re gonna get three outside the brackets and then 𝑥 minus two inside the
brackets. And that’s because if we have three
multiplied by 𝑥 gives us three 𝑥 and three multiplied by negative two gives us
negative six.

So at this point, we give a quick
tip. You should have the same factor in
each bracket. If you haven’t for some reason, so
if you didn’t have 𝑥 minus two in each of them, then what you would do is look back
at your working out because something would have gone amiss, because if you don’t
have that, we won’t be able to split it into its two factors.

So therefore, we can say fully
factorised two 𝑥 squared minus 𝑥 minus six equals zero gives us two 𝑥 plus three
multiplied by 𝑥 minus two is equal to zero. So therefore, we can say that two
𝑥 plus three multiplied by 𝑥 minus two is gonna be greater than zero. So that’s our inequality.

What we need to do now is actually
find our roots, so where is it actually going to meet the 𝑥-axis. And to do that, what we do is we
actually make each bracket equal to zero. And the reason we do that is
because if we have zero multiplied by any number, then that’s gonna give us an
answer of zero. So it will give us our roots to
tell us where we’re actually gonna cross the 𝑥-axis.

So we start off with two 𝑥 plus
three equals zero. So therefore, if we subtract three
from each side, we’re gonna get two 𝑥 equals negative three. And then if we actually divide each
side by two, we get 𝑥 is equal to negative three over two. So that’s our first root.

Our second root we’ll find by
making the next bracket equal to zero. So we get 𝑥 minus two is equal to
zero. So therefore, if we add two to each
side of the equation, we get 𝑥 is equal to two. So this is our other root.

So now what I’ve done is I’ve drawn
a rough sketch of actually our graph. And the reason we do that is we
need to show in every type of this question how we actually come about the
interval. So if we’re looking for an interval
to find the solution to an inequality, we can’t just write the answer down. We have to show where it’s come
from. The key here is that we’re actually
looking for the points where our inequalities are actually greater than zero.

Now we know that it’s a U-shaped
parabola, because actually we’ve got a positive 𝑥 squared term. So we’re ok with the shape. We’ve also got another two
roots. Which part of the graph though are
we interested in? Well, we’re actually interested in
the positive parts of our graph, so therefore everything above the 𝑥-axis.

So therefore, what we’re actually
looking for is any 𝑥 value that is less than negative three over two or greater
than two. So therefore, we can say that our
𝑥 value is gonna be 𝑥 is greater than two or 𝑥 is less than negative three over
two.

But is this correct? Do we want both answers? Well, the key here is the word
“possible,” cause in the question it says find all possible values of 𝑥. And actually, one of these would
not be possible. So let’s consider our second
inequality that we had, which was 𝑥 is less than negative three over two.

Okay, I’m gonna pick a value of 𝑥
which is less than negative three over two. I’m gonna pick a value which is
negative two. So 𝑥 is negative two. If we substitute that in for the
dimensions of our perpendicular height of a parallelogram, we’re gonna get negative
two minus one, which would actually give us a height of negative three. Well, this can’t be true. We can’t have a height of negative
three. So therefore, we can say that
actually that isn’t valid.

So therefore, we can say that if
the area of the parallelogram is greater than the area of the rectangle, the
possible values of 𝑥 are 𝑥 is greater than two.