Video: Derive the Quadratic Formula | Nagwa Video: Derive the Quadratic Formula | Nagwa

# Video: Derive the Quadratic Formula

Here, we will explain how to use the technique of completing the square to derive a formula to solve an equation of the general form ππ₯Β² + ππ₯ + π = 0 for π₯.

09:27

### Video Transcript

Find a formula for solving a quadratic equation of the form ππ₯ squared plus ππ₯ plus π is equal to zero. So hereβs our equation weβre trying to solve in terms of π₯, and π, π, and π are just constant numbers.

So first of all, weβre gonna divide both sides of the equation by the constant π. So the πβs here cancel out and weβre left with ππ₯ over π plus π over π is equal to zero over π. Well thatβs just zero. So with just a little bit of tidying up, this is the equation weβre now trying to solve.

And now Iβm gonna subtract this π over π from both sides of my equation, so subtracting from the left-hand side and subtracting from the right-hand side. So π over π take away π over π is nothing, and on the right-hand side at zero take away π over π, Iβve just got negative π over π.

Now Iβm gonna use the method of completing the square on the left-hand side to try and find another form of that left-hand side, which is gonna help us to solve our equation. So this is our first guess: what weβre going to square in order to get an equivalent expression to this one.

Now the π₯ squared term tells us weβre gonna use π₯ here and then we take half of this coefficient for this term here. So letβs multiply out these brackets. π₯ times π₯ is π₯ squared, π₯ lots of π over two π is π over two ππ₯, π over two π lots of π₯ is another π over two ππ₯, and finally π over two π times π over two π is positive π squared over four π squared.

Then adding π over two ππ₯ plus π over two ππ₯ gives us π over ππ₯. Now the first part here π₯ squared plus π over ππ₯ is exactly what we were looking for up here, π₯ squared over πππ₯. The problem is, weβve got this extra bit on the end, the extra π squared over four π squared.

So this expression here isnβt the same as this expression here. Weβve got an extra π squared over four π squared. So if I took that away from this term, then I would end up with what I wanted at the very beginning.

So our first guess was π₯ plus π over two π all squared would be equivalent to π₯ squared plus π over ππ₯, but as weβve seen down here, thatβs not quite right. It gives us a little bit too much, so I need to subtract that from this in order to get what I was actually looking for. So hopefully you can see that these two lines are the same thing; π₯ plus π over two π all squared gives us this expression down here.

If I subtract π squared from four π squared, thatβs like getting rid of that bit, which just leaves this bit here, which is in fact exactly what we were looking for. Right so now Iβm gonna try and isolate the π₯ plus π over two π all squared term by adding π squared over four π squared to both sides of my equation.

Now we can see that, on the left-hand side, Iβve got negative π squared over four π squared, and then Iβm adding π squared over four π squared, so those two terms are gonna cancel out. And now Iβm gonna try and manipulate the right-hand side so that, instead of two separate terms, Iβve just got one big term.

And the way I can do this, Iβve got a denominator on this term of four π squared. Iβve got a denominator on this term of just π. So if I multiply the top by four π and the bottom by four π, now my denominator on this term is gonna be four π squared just like this term.

Now by multiplying by four π over four π, that basically is one, four π divided by four π is one, so I havenβt changed the magnitude of the number; Iβve just changed the format of the expression. So when I rewrite that, you can see that Iβve got two terms with a common denominator, so I can just put that altogether in one big term.

And there we have it. So rather than write minus four ππ plus π squared, Iβve put the positive term first, so π squared minus four ππ all over four π squared. Now I can take square roots of both sides so that I can see on the left-hand side Iβm just gonna find out what π₯ plus π over two π is.

And clearly there are two possible answers for the right-hand side. We could have the positive version of the square root of π squared minus four ππ all over four π squared or we could have the negative version of that. And now what Iβm gonna do as well is Iβm just gonna rearrange that second term slightly because the square root of four π squared is clearly two π.

And now I can simply subtract π over two π from both sides, so that I can leave π₯ on its own over here on the left. So weβre going to subtract π over two π from both sides of the equation. So over here on the left-hand side, Iβve got π over two π, take away π over two π, so that leaves nothing, so those two things cancel out.

And now Iβm gonna write this in a slightly different order, so Iβm gonna write the negative π over two π first. And weβve already got a common denominator here, so in fact Iβm gonna go one stage further than that. Iβm gonna write this all as one expression on the right-hand side.

And here we have a general formula for solving quadratic equations of the format ππ₯ squared plus ππ₯ plus π is equal to zero. So letβs see how we can use that in practice. Hereβs a question. Solve π₯ squared minus five π₯ plus six equals zero. So weβve gotta find the value or values of π₯ which will satisfy that equation.

So the first thing to do is to work out what π, π, and π are from the general formula. So our general format is ππ₯ squared plus ππ₯ plus π, so the coefficient of π₯ squared is the value of π. Thereβs nothing in front of the π₯ squared, so that means thereβs one π₯ squared, so π is equal to one.

The coefficient in front of the π₯ is negative five, so π is negative five. And the number term on its own at the end is positive six, so π is positive six. So next we write out our quadratic formula: π₯ is equal to negative π plus or minus the square root of π squared minus four ππ all over two π.

And then weβre gonna substitute in these values to that equation. Now a couple of top tips: I would always write the equation out in full as we have done here substituting in all values π, π, and π exactly as they are. That shows if youβre in exam it shows the examiner you know basically what youβre doing.

Secondly I would always write negative numbers in brackets, so weβve got negative five. Iβve written the negative five in brackets here and here because π was negative five. This means that when you enter the number on your calculator, your calculator isnβt going to make any mistakes. So for example here, negative five squared if Iβve just written negative five squared on my calculator that will give me an answer of negative twenty-five.

But really we want the whole of negative five squared, so negative five times negative five is positive twenty-five. So this is the answer weβre looking for. Therefore, the brackets are definitely gonna help us get the right answer. Okay letβs go through that bit by bit.

The negative of negative five is positive five plus or minus negative five times negative five is positive twenty-five, so this is gonna be twenty-five, and four times one is four and four times six is twenty-four, so weβre gonna be taking away twenty-four. And on the bottom, two times one is just two.

So working out the contents of the square root, there twenty-five take away twenty-four is one, so weβve got the square root of one. So the- the answers are gonna be five plus the square root of one all over two or five take away the square root of one all over two. So now we write those out separately.

First of all, weβre gonna do the positive case, then weβre going to do the negative case. So the first instance π₯ could be five plus one over two, which gives us an answer of three, or π₯ could be five take away one over two, which gives us an answer of two. Now we just need to write our answer out clearly.

There we go; the answer π₯ equals three or π₯ equals two. If we wanted to be really sure of ourselves, we could always try those back in the original equation, so letβs just quickly do that now. So three, letβs try three first, three squared is nine take away five lots of three, so weβre taking away fifteen, and then add six, so nine add six is fifteen takeaway fifteen, thatβs zero, thatβs correct.

And now with two, two squared is four take away five lots of two, so weβre taking away ten, and then weβre adding six, well four plus six is ten take away ten is equal to zero. So we know thatβs correct as well, so we- weβre completely happy that our answers are correct.

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