### Video Transcript

Solve 𝑥 squared plus three 𝑥
minus six equals zero. So exactly the same technique; it’s
just that towards the end of this question the numbers are gonna get a little bit
trickier. But never mind we’ll work out how
to do that when we get there. So we’ll clear off a little space
on the left-hand side by adding six to both sides and then we’ll complete the square
on the left-hand side. So we’ve got 𝑥 and we take half of
the 𝑥 coefficient, three over two; we’re squaring that. To make that equivalent to the
expression above it, we’re gonna need to take away that three over two term all
squared. So that’s equal to six. So I’m just gonna evaluate the
three over two all squared. So I’ve got 𝑥 plus three over two
squared minus nine over four is equal to six and I’m gonna add the nine over four to
both sides. So I’ve got six plus nine over four
on the right-hand side. So I need to find an equivalent
version of six, which is actually a top heavy fraction over four. So I’ve got my common
denominators. And that’s obviously twenty-four
over four because twenty-four over four is the same as six. Now twenty-four over four plus nine
over four equals thirty-three over four. So I’ve got 𝑥 plus three over two
squared is equal to thirty-three over four. Now I’m gonna take square roots of
both sides. And the square root of the
left-hand side is just 𝑥 plus three over two and we got two possible values for the
right-hand side: it’s either the positive version of the square root of thirty-three
over four or it’s the negative version. Now obviously that’s the same as
root thirty-three over root four. So that’s the root of four is
two.

So this is slightly different to
the previous questions because always before when we were doing this square rooting
stage, we ended up with nice simple numbers. But the square root of thirty-three
isn’t anything nice and easy. So we just leave it as root
thirty-three. And in the next step to get 𝑥 on
its own, on the left-hand side here I’m subtracting three over two from both
sides. And so the right hand becomes
negative three over two plus or minus root thirty-three over two. So as luck would have it here, I’ve
already got my common denominator. So I’ve now got two possible values
for 𝑥: it could be negative three over two plus the root of thirty-three over two
or it could be negative three over two subtract this root of thirty-three over
two. Now I could gather those terms
together but because these numbers are not particularly nice, I have to leave them
in-in that sort of format. Now the question might-might have
asked me to give my answers rounded correct to one decimal place or it might have
asked me to leave them in this exact format. And just for the sake of argument,
I’m gonna pretend they asked me to leave it in the exact format in this case. So sometimes the answers work out
really simply and sometimes the numbers are a little bit more complicated, but the
method is just the same.