### 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.