### Video Transcript

In this video, we’re gonna see how to use the quadratic formula to solve a quadratic equation, or find its roots. We’ll work through a few examples and explain how to avoid some of the common mistakes that people make. We’ll look at a range of simpler questions to see how to use the formula, and then one or two slightly trickier ones, we have to rearrange the equation and one that even breaks the formula.

So our first example is: Solve two 𝑥 squared minus four 𝑥 minus thirty equals zero.

So we’re gonna use the quadratic formula to solve this quadratic equation. And the quadratic formula is, if 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero, provided 𝑎 is not equal to zero, because if 𝑎 was equal to zero, then it wouldn’t be a quadratic because the coefficient of 𝑥 squared will be zero, then 𝑥 is equal to the negative of 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. So what we need to do is look at our equation and match up the 𝑎, 𝑏, and 𝑐, and then just plug them into that formula. And the multiple of 𝑥 squared is two, so 𝑎 is two. The multiple of 𝑥, or the coefficient of 𝑥, is negative four. So 𝑏 is negative four. And the constant term on its own is negative thirty, so 𝑐 is equal to negative thirty. So luckily for us, our equation was already in the format something times 𝑥 squared plus something times 𝑥 plus a number is equal to zero, so we didn’t have to do any rearranging. Now we’ve worked out what 𝑎, 𝑏, and 𝑐 are. We can just plug them into this formula.

Now my first top tip is to make sure you write it all out in full. So plug all the numbers in, and write the full formula out. And the second top tip is to put all of your negative numbers in brackets. So 𝑏 was negative four, so over here we’ve put that in brackets, so the negative of negative four. In this term here, up there, we’ve got negative four all squared. That means it’s negative four times negative four. And over here we’ve got 𝑐 is negative thirty. So again, I’ve just put that in brackets so that we know that we’re multiplying by negative thirty and not just subtracting thirty from the end of that calculation. Now if 𝑏 is negative, this is especially important because if you just type negative four squared into your calculator, you’ll get the answer negative sixteen, and that’s not quite what we’re actually looking for.

So logic on your calculator probably does this squared, before it does the negative. And that means that it ends up with four squared is sixteen, and then it takes the negative there, to give you negative sixteen. So your calculator is interpreting the calculation like this: The negative of four times four, whereas what we really wanted was negative four times negative four. So to force your calculator to do the correct calculation, make sure you put your negative numbers in brackets.

So let’s evaluate some of this now. Well the negative of negative four is positive four. And then we’ve got to add or subtract, we’ll talk about that even more in a moment. And then in the brackets negative four times negative four, as we’ve just said, is positive sixteen. And then we’ve got four times two times negative thirty. Well that’s negative two hundred and forty. So we’re taking away negative two hundred and forty, which means we’re adding two hundred and forty. Now that’s all over two times two, which is four. And then sixteen plus two hundred and forty is two hundred and fifty-six. And the square root of two hundred and fifty-six is exactly sixteen. So we’ve got four plus or minus sixteen all over four. Now this is where the plus or minus comes into play, this means we’ve got two calculations to do. We’ve either got 𝑥 is equal to four plus sixteen all over four or 𝑥 is equal to four minus sixteen all over four. And four plus sixteen is twenty, so 𝑥 could be twenty over four or 𝑥 could be four minus sixteen is minus twelve, so that’s minus twelve over four.

So we’ve got two answers: 𝑥 is equal to five or 𝑥 is equal to negative three. That means that there are two values of 𝑥, that if I plug them in here and here in my original equation, I will make that equation true. So let’s just quickly check that then. So if I plug in 𝑥 equals five, that means I’ve got two times five squared minus four times five minus thirty. Well five squared is twenty-five, so that’s two times twenty-five. And four times five is twenty. So I’m taking away twenty, and then I’m taking away another thirty. Well two times twenty-five is fifty. And if I do fifty take away twenty take away thirty, I get zero, and that was what I was looking for. So let’s try again with 𝑥 is negative three.

So plugging negative three in there, I’ve got two lots of negative three squared minus four lots of negative three, and then I’m taking away another thirty. Well negative three squared, negative three times negative three is positive nine. So that first term becomes two times nine. Then I’m taking away four lots of negative three, well four lots of negative three is negative twelve. So if I’m taking away negative twelve, that means I’m adding twelve. And then we’re taking away thirty and two times nine is eighteen. So I’ve got eighteen plus twelve minus thirty, and that of course is zero, which again is what we were looking for. So that’s great! Looks like we’ve got the right answers.

Now that is the end of that question, but it’s worth mentioning the fact that because we had whole number answers, five and negative three, we’ve got integer answers, that means that the original expression here would’ve factored. So in fact, if I factor two 𝑥 squared minus four 𝑥 minus thirty, I can do that as two 𝑥 plus six times 𝑥 minus five. Because two 𝑥 times 𝑥 is two 𝑥 squared, two 𝑥 times negative five is negative ten 𝑥, six times 𝑥 is positive six 𝑥, and six times negative five is negative thirty. Negative ten 𝑥 plus six 𝑥 is negative four 𝑥. So yes, when I multiply out two 𝑥 plus six times 𝑥 minus five, I do get the expression I was looking for two 𝑥 squared minus four 𝑥 minus thirty. And that means I’ve got something times something is equal to zero. And anyway, something times something could be equal to zero if one of those is zero. So either this is equal to zero or this is equal to zero. And solving both of those, I get the same answers, 𝑥 is negative three or 𝑥 is equal to five.

Now we’re going way beyond the call of duty here and we’re drawing out the graph as well, which we don’t need to do. But if I do draw out the graph of 𝑦 equals two 𝑥 squared minus four 𝑥 minus thirty, I can see that the 𝑦-coordinate is zero here and here, and that’s at negative three, when 𝑥 is negative three, or when 𝑥 is equal to five. And they’re the answers that we’ve got as well. So we’re just trying to visualise what that answer looks like on the graph.

Okay. Let’s move on to our second example then. Solve seven 𝑥 squared plus twelve 𝑥 plus two equals zero. So this time, 𝑎 is seven, 𝑏 is twelve, and 𝑐 is two. They’re all positive. So I can put those into the formula. And then I get negative twelve plus or minus the square root of twelve squared minus four times seven times two all over two times seven, just by plugging in 𝑎, 𝑏 and 𝑐 into that formula. Well twelve squared is a hundred and forty-four, and four times seven is twenty-eight, and twenty-eight times two is fifty-six, so we’re subtracting fifty-six from that.

And again, we’ve got two values. Either 𝑥 is equal to negative twelve plus root eighty-eight all over fourteen or 𝑥 is equal to negative twelve minus root eighty-eight all over fourteen. Now root eighty-eight turns out to be an irrational number. We can’t just find a nice easy integer square root for it. So typically, in these questions you would either give your answer in its simplest surd form, or you would round to a specific number of decimal places. So either 𝑥 is equal to negative six plus root twenty-two over seven, which is minus nought point one nine to two decimal places, or 𝑥 is equal to negative six take away root twenty-two all over seven which is equal to minus one point five three to two decimal places. So that was probably a more typical example of the sort of question you’d be asked to answer, where the original expression up here didn’t factor nicely, so we couldn’t come up with nice simple answers.

Number there then asks us to solve two 𝑥 squared minus five is equal to negative six 𝑥. But remember, our quadratic formula requires the equation to be in this format, 𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. So what I need to do is add six 𝑥 to both sides and then I’ve got something times 𝑥 squared plus something times 𝑥 plus a number is equal to zero. And that gives me two 𝑥 squared plus six 𝑥 minus five is equal to zero. So now 𝑎 is equal to two, 𝑏 is equal to six, and 𝑐 is equal to negative five. So I can just plug those into my formula, which gives me negative six plus or minus the square root of six squared minus four times two times negative five all over two times two. So simplifying those numbers, six squared is thirty-six, and four times two is eight times five is forty. So that was negative forty. We’re taking away negative forty, which means we’re adding forty. So that means 𝑥 is equal to negative six plus or minus root seventy-six all over four. So that gives us two calculations, one of them is negative six plus root seventy-six all over four, and the other is 𝑥 is equal to negative six take away root seventy-six all over four.

So again I’ve got two different answers, 𝑥 is equal to negative three plus root nineteen over two which is nought point six eight to two decimal places, or 𝑥 is equal to negative three take away root nineteen over two, and that to two decimal places is negative three point six eight.

So number four was asked in a slightly different format. Rather than asking you to solve two 𝑥 squared plus four 𝑥 plus two equals zero, you’re being asked to find the roots of 𝑦 equals two 𝑥 squared plus four 𝑥 plus two. It’s the same question, to find the roots of something, you’ve just got to find what 𝑥-coordinate generates a 𝑦-coordinate of zero. And when we put that equal to zero, we’ve got all positives for 𝑎, 𝑏, and 𝑐. 𝑎 is two, 𝑏 is four and 𝑐 is two. So plugging those into the quadratic formula, we get 𝑥 is equal to negative four plus or minus the square root of four squared minus four times two times two all over two times two. So just evaluating those, well four squared is sixteen and four times two is eight times two is sixteen. So we’re taking away sixteen and that’s all over four. Well obviously sixteen take away sixteen is nothing. So we’ve got negative four plus or minus the square root of nothing all over four. Well of course negative four add nothing and negative four take away nothing, gives you just negative four. So in both cases, we’ve got negative four over four, which is negative one.

So in this particular case, the way the numbers, the 𝑎s, the 𝑏s, the 𝑐s have worked out in that formula, we’ve got two answers but they’re both exactly the same. And this is called repeated roots. So in fact, we can just express that 𝑥 equals negative one is the only value of 𝑥, which is ge- generate a 𝑦-coordinate of zero in that original equation up here.

And in fact, just for the sake of interest, if we go back and look at what the graph would look like, that’s what it looks like. And we can see here that there is only one place that that touches the 𝑥-axis that that has effectively a 𝑦-coordinate of zero, and that is at negative one.

Okay. Let’s look at number five then. Solve three 𝑥 squared plus two 𝑥 plus four is equal to zero. So if we just write out our quadratic formula there, we can see that 𝑎 is three, 𝑏 is two, and 𝑐 is four. So plugging those values for 𝑎, 𝑏, and 𝑐 into the quadratic formula gives us 𝑥 is equal to negative two plus or minus the square root of two squared minus four times three times four all over two times three. Well, two squared is four, four times three times four is forty-eight, so we’ve got four take away forty-eight in that square root. And then two times three on the denominator is six. So two possible solutions, negative two plus the square root of negative forty-four all over six, or 𝑥 is equal to negative two minus the square root of negative forty-four all over six. But if you try typing that into your calculator, you’ll get a math error. The problem is this bit here, square root of negative forty-four. There aren’t any real numbers that you can multiply by themselves to give a negative answer. If you take a negative number and multiply it by itself, you get a positive answer. And if you take a positive number and you multiply it by itself, you get a positive answer. So there isn’t a real number that you can multiply by itself that will give a negative answer.

Now if we have a look at the graph of 𝑦 equals three 𝑥 squared plus two 𝑥 plus four, and then we put 𝑦 equal to zero. We can see that, in fact, there aren’t any 𝑥-values that are gonna generate a 𝑦-coordinate of zero. That curve doesn’t cut through the 𝑥-axis here. There are no 𝑥-values that generate a 𝑦-coordinate of zero.

So this was a bit of a trick question that broke the formula. We were asked to solve something that doesn’t have any real solutions. So it seems like a bit of a trick question, but really it’s just saying, you know, where does this quadratic curve cut the 𝑥-axis on a curve that doesn’t cut the 𝑥-axis. So that’s what we found out. So when you’ve got a negative value here for 𝑏 squared minus four 𝑎𝑐, you know you’ve got a curve that doesn’t cut the 𝑥-axis.

Right. Just to see how well you’ve been paying attention, now I’m gonna set you a question, number six, and I want you to have a go at doing this. Here is some working out where a student has correctly substituted values into the quadratic formula, which we’ve been given here. What is the quadratic equation they were trying to solve? So it is something that was equal to zero. Now their working out says 𝑥 is equal to five plus or minus the square root of twenty-five minus eighty all over eight. So what I like you to do is pause the video and then have a go at this question and see if you can work out what you think the equation was.

Well looking at what they’ve written, we can see that on the denominator it should be two 𝑎, so eight is equal to two 𝑎. And so, if I divide both sides of that equation by two, it seems that the value of 𝑎 would’ve been four. Now also the value of negative 𝑏 turns out to be five. So if I multiply both sides of that equation by negative one, I’ve got 𝑏 is equal to negative five. And lastly, we can see that four times 𝑎 times 𝑐 is equal to eighty. But we know that 𝑎 was equal to four, so we can put that value in here. So that means four times four times 𝑐 is equal to eighty, or sixteen times 𝑐 is equal to eighty. So if I divide both sides of that equation by sixteen, sixteen 𝑐 divided by sixteen is just 𝑐, and eighty divided by sixteen is five. So 𝑎 was four, 𝑏 was negative five, and 𝑐 was five. Now they’ve said what is the quadratic equation they were trying to solve. So we can put those into this equation up here, and we know that the equation they were trying to solve was four 𝑥 squared minus five 𝑥 plus five is equal to zero.

Well, you’ve seen a range of different quadratic formula-type questions there. So hopefully, you’re now ready to go and tackle any quadratic formula-type question that you may happen to come across.