WEBVTT
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Find a formula for solving a quadratic equation of the form ππ₯ squared plus ππ₯ plus π is equal to zero.
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So hereβs our equation weβre trying to solve in terms of π₯, and π, π, and π are just constant numbers.
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So first of all, weβre gonna divide both sides of the equation by the constant π.
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So the πβs here cancel out and weβre left with ππ₯ over π plus π over π is equal to zero over π.
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Well thatβs just zero.
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So with just a little bit of tidying up, this is the equation weβre now trying to solve.
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And now Iβm 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.
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So π over π take away π over π is nothing, and on the right-hand side at zero take away π over π, Iβve just got negative π over π.
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Now Iβm 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.
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So this is our first guess: what weβre going to square in order to get an equivalent expression to this one.
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Now the π₯ squared term tells us weβre gonna use π₯ here and then we take half of this coefficient for this term here.
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So letβs multiply out these brackets.
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π₯ times π₯ is π₯ squared, π₯ lots of π over two π is π over two ππ₯, π over two π lots of π₯ is another π over two ππ₯, and finally π over two π times π over two π is positive π squared over four π squared.
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Then adding π over two ππ₯ plus π over two ππ₯ gives us π over ππ₯.
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Now the first part here π₯ squared plus π over ππ₯ is exactly what we were looking for up here, π₯ squared over πππ₯.
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The problem is, weβve got this extra bit on the end, the extra π squared over four π squared.
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So this expression here isnβt the same as this expression here.
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Weβve got an extra π squared over four π squared.
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So if I took that away from this term, then I would end up with what I wanted at the very beginning.
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So our first guess was π₯ plus π over two π all squared would be equivalent to π₯ squared plus π over ππ₯, but as weβve seen down here, thatβs not quite right.
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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.
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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.
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If I subtract π squared from four π squared, thatβs like getting rid of that bit, which just leaves this bit here, which is in fact exactly what we were looking for.
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Right so now Iβm 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.
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Now we can see that, on the left-hand side, Iβve got negative π squared over four π squared, and then Iβm adding π squared over four π squared, so those two terms are gonna cancel out.
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And now Iβm gonna try and manipulate the right-hand side so that, instead of two separate terms, Iβve just got one big term.
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And the way I can do this, Iβve got a denominator on this term of four π squared.
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Iβve got a denominator on this term of just π.
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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.
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Now by multiplying by four π over four π, that basically is one, four π divided by four π is one, so I havenβt changed the magnitude of the number; Iβve just changed the format of the expression.
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So when I rewrite that, you can see that Iβve got two terms with a common denominator, so I can just put that altogether in one big term.
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And there we have it.
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So rather than write minus four ππ plus π squared, Iβve put the positive term first, so π squared minus four ππ all over four π squared.
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Now I can take square roots of both sides so that I can see on the left-hand side Iβm just gonna find out what π₯ plus π over two π is.
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And clearly there are two possible answers for the right-hand side.
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We could have the positive version of the square root of π squared minus four ππ all over four π squared or we could have the negative version of that.
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And now what Iβm gonna do as well is Iβm just gonna rearrange that second term slightly because the square root of four π squared is clearly two π.
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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.
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So weβre going to subtract π over two π from both sides of the equation.
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So over here on the left-hand side, Iβve got π over two π, take away π over two π, so that leaves nothing, so those two things cancel out.
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And now Iβm gonna write this in a slightly different order, so Iβm gonna write the negative π over two π first.
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And weβve already got a common denominator here, so in fact Iβm gonna go one stage further than that.
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Iβm gonna write this all as one expression on the right-hand side.
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And here we have a general formula for solving quadratic equations of the format ππ₯ squared plus ππ₯ plus π is equal to zero.
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So letβs see how we can use that in practice.
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Hereβs a question.
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Solve π₯ squared minus five π₯ plus six equals zero.
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So weβve gotta find the value or values of π₯ which will satisfy that equation.
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So the first thing to do is to work out what π, π, and π are from the general formula.
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So our general format is ππ₯ squared plus ππ₯ plus π, so the coefficient of π₯ squared is the value of π.
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Thereβs nothing in front of the π₯ squared, so that means thereβs one π₯ squared, so π is equal to one.
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The coefficient in front of the π₯ is negative five, so π is negative five.
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And the number term on its own at the end is positive six, so π is positive six.
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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 ππ all over two π.
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And then weβre gonna substitute in these values to that equation.
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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.
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That shows if youβre in exam it shows the examiner you know basically what youβre doing.
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Secondly I would always write negative numbers in brackets, so weβve got negative five.
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Iβve written the negative five in brackets here and here because π was negative five.
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This means that when you enter the number on your calculator, your calculator isnβt going to make any mistakes.
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So for example here, negative five squared if Iβve just written negative five squared on my calculator that will give me an answer of negative twenty-five.
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But really we want the whole of negative five squared, so negative five times negative five is positive twenty-five.
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So this is the answer weβre looking for.
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Therefore, the brackets are definitely gonna help us get the right answer.
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Okay letβs go through that bit by bit.
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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 weβre gonna be taking away twenty-four.
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And on the bottom, two times one is just two.
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So working out the contents of the square root, there twenty-five take away twenty-four is one, so weβve got the square root of one.
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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.
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So now we write those out separately.
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First of all, weβre gonna do the positive case, then weβre going to do the negative case.
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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.
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Now we just need to write our answer out clearly.
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There we go; the answer π₯ equals three or π₯ equals two.
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If we wanted to be really sure of ourselves, we could always try those back in the original equation, so letβs just quickly do that now.
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So three, letβs try three first, three squared is nine take away five lots of three, so weβre taking away fifteen, and then add six, so nine add six is fifteen takeaway fifteen, thatβs zero, thatβs correct.
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And now with two, two squared is four take away five lots of two, so weβre taking away ten, and then weβre adding six, well four plus six is ten take away ten is equal to zero.
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So we know thatβs correct as well, so we- weβre completely happy that our answers are correct.