Video Transcript
Find all the real solutions to the
system of equations 𝑦 equals negative two 𝑥 squared plus four 𝑥 plus six and 𝑦
equals four 𝑥 squared plus 22𝑥 plus 24.
So when we take a look at our
system of equations, we can see that they both are 𝑦 equals and then we have an
expression. So therefore, what we can do to
solve the problem is equate them to each other. So therefore, what we can say is
that negative two 𝑥 squared plus four 𝑥 plus six is equal to four 𝑥 squared plus
22𝑥 plus 24.
So now what we want to do is we
want to actually have a quadratic in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals
zero to help us solve it. So in order to do that, what we’re
gonna do is add two 𝑥 squared, subtract four 𝑥, and subtract six from both sides
of the equation. And when we do that, what we’re
gonna be left with is zero is equal to six 𝑥 squared plus 18𝑥 plus 18. And as we can see that each term on
the right-hand side is divisible by six, what we can do is divide through by six to
make it easier to solve. So when we do that, what we’re
gonna get is zero is equal to 𝑥 squared plus three 𝑥 plus three.
Or straightaway, we can identify
that this is in fact not going to factor, because we could factor to solve the
quadratic but in this case it won’t work because we have three. And we need to have a product that
makes three. Well the only product that makes
three is one multiplied by three, which equals three. Well, if you add one and three, you
get four, whereas we want to have a sum of three. So we know that it’s not going to
factor.
Well, we’re now at the stage where
we think how can we solve our quadratic using other methods? What we could do is plug it into
the quadratic equation and have a look at how we get on. And we will do that to show how we
get the same result.
However, what you might notice by
inspection is that the coefficient of 𝑥 and our numerical term are both positive
and the same value and less than five. And also, it’s worth noting, which
is important, the 𝑥 squared coefficient is also positive. And by noticing this, we can
actually think, “Well, is it possible that there couldn’t be any real roots at
all?”
And what we can do to check this
out is use the discriminant, because what we know from the discriminant is that if
we have 𝑏 squared minus four 𝑎𝑐 is less than zero, then there are no real
roots. If 𝑏 squared minus four 𝑎𝑐 is
equal to zero, then there’s one real root. And if we have 𝑏 squared minus
four 𝑎𝑐 is greater than zero, then there are two distinct real roots.
And in our equation, 𝑎 is equal to
one because that’s the coefficient of 𝑥 squared. 𝑏 is equal to three because it’s
our coefficient of 𝑥. And then our numerical term 𝑐 is
equal to three as well.
So now what we can do is work out
our discriminant. And to do that, we’ll have three
squared minus four multiplied by one multiplied by three. So our discriminant’s gonna be nine
minus 12. So our discriminant is gonna be
negative three. So the discriminant therefore is
less than zero. So we can say that they’re gonna be
no real roots.
And we managed to notice this
because, as I said, we had the coefficient of 𝑥 was positive three and the
numerical value was positive three. And we also had the coefficient of
𝑥 squared being positive. So therefore, what we know is that
it was going to always look like they were gonna be no real roots. Because if we think about it, if we
have a number squared and then we subtract four multiplied by that number and
another number, it’s always gonna be a negative value.
Now this was using the
discriminant. What I did mention was that we
could have found this out also if we’d just gone straight into using the quadratic
formula. Well, the quadratic formula is that
𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four
𝑎𝑐 over two 𝑎. Well, if we use the quadratic
formula, what we’d have is 𝑥 equals negative three plus or minus the square root of
three squared minus four multiplied by one multiplied by three all over two
multiplied by one.
So then this would give us 𝑥 is
equal to negative three plus or minus the square root of negative three over
two. So now what we get is root negative
three in our numerator. Well, root negative three is
undefined, because we can’t have a result for the square root of a negative number
if we’re using real numbers. So therefore, what we could say is
that the system of equations would have no real roots.
