Video: Pack 1 • Paper 1 • Question 20

Pack 1 • Paper 1 • Question 20

07:49

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.

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