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Using the method of systematic trial and improvement and giving your answer correct to one decimal place, find the solution between đť‘Ą equals zero and đť‘Ą equals one to two đť‘Ą squared minus five đť‘Ą plus one equals zero.
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So theyâ€™re asking for the particular method, systematic trial and improvement.
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Theyâ€™ve given you the level of accuracy that we need, correct to one decimal place.
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And theyâ€™ve given us some starting conditions here, we want to find the solution between đť‘Ą equals zero and đť‘Ą equals one.
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And theyâ€™ve given us an equation to solve.
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Now normally, the first thing you do with these questions is to draw a table.
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But what weâ€™re gonna do is just have a quick look at the graph.
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Now obviously you wouldnâ€™t normally see the graph, but thatâ€™s gonna help us to see how this method is gonna work.
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So hereâ€™s the graph of đť‘¦ equals two đť‘Ą squared minus five đť‘Ą plus one.
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Now weâ€™re looking for the specific solution where that is equal to zero.
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In other words, where the đť‘¦-coordinate here is equal to zero.
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And when the đť‘¦-coordinate is zero, weâ€™re talking about any point on the đť‘Ą-axis.
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Now this graph cuts the đť‘Ą-axis in two places, here and here.
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And the question has said that weâ€™re looking for this solution here between đť‘Ą equals zero and đť‘Ą equals one.
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So this is the one that weâ€™re looking for.
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So weâ€™re trying to find the đť‘Ą-coordinate of this point: the value of đť‘Ą which generates a đť‘¦-coordinate of zero.
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So zooming in on that curve between đť‘Ą is zero and đť‘Ą is one, what weâ€™re gonna do is use the đť‘Ą-coordinate of zero and plug that into the equation and see what value the expression has.
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So two times zero squared minus five times zero plus one.
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And that hopefully is gonna give us an answer of one.
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Weâ€™re then gonna do the same for đť‘Ą equals one to see what expression value is generated.
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And weâ€™ll see that one is too big and one is too small.
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So our solution that generates an answer of zero must be somewhere between đť‘Ą equals zero and đť‘Ą equals one.
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Here, itâ€™s too big; itâ€™s getting smaller and smaller and smaller.
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Somewhere it becomes zero and then it gets smaller and smaller; itâ€™s negative and goes down here.
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Now obviously we wonâ€™t have the graph to look at.
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But what weâ€™ll be able to do is, once we know that one is too small and one is too big, we will be able to pick a point somewhere in the middle here.
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And weâ€™ll be able to calculate the new expression value.
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And weâ€™ll have a new value that gives us an answer thatâ€™s too small and a new value that gives us an answer thatâ€™s too big.
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And now- then weâ€™ll try to look between those two points and weâ€™ll keep moving that đť‘Ą-value in either from the right or from the left.
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And weâ€™ll gradually zoom in on our actual đť‘Ą-value that weâ€™re looking for here.
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So letâ€™s go ahead and try that.
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So weâ€™ll construct the table thatâ€™s got these four columns.
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First of all, weâ€™ve got the đť‘Ą-values that weâ€™re gonna try.
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And weâ€™ve already been told what values to use for our first two attempts, zero and one.
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The second column is where weâ€™re gonna plug those đť‘Ą values into the expression we were given in the equation and evaluate that expression.
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So with zero then, weâ€™ve got two lots of zero squared minus five lots of zero plus one.
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And two times zero squared is zero, take away five lots of zero, thatâ€™s zero, plus one leaves us one.
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Then weâ€™re gonna try plugging in đť‘Ą equals one.
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So two times one squared minus five times one plus one.
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And I give this an answer of negative two.
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So now, we can fill out the comments.
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And when đť‘Ą was zero, the result was one; thatâ€™s too big.
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We wanted our result to be zero, remember, because thatâ€™s what we had in the original equation.
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And in the second case, the result was too small because it was less than zero.
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So our quadratic curve is gonna be a smooth curve.
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So somehow itâ€™s gonna go between this point and this point; there must be an đť‘Ą-value in between zero and one which will generate a đť‘¦-coordinate of zero when I plug it in.
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So I can now fill in the â€śRangeâ€ť column.
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And I know from those two results â€” one being too big one being too small â€” that somewhere between the two there must be a value of đť‘Ą which generates an đť‘Ą- a đť‘¦-coordinate of zero.
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Now itâ€™s up to you which particular value you choose.
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Iâ€™m gonna go and try an đť‘Ą-value of nought point three.
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And when I do that, I try plugging nought point three into that expression.
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So two times nought point three squared minus five times nought point three plus one.
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And that generates a result of nought point three two, a negative nought point three two.
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So thatâ€™s below zero.
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So thatâ€™s too small as well.
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So just drawing that on my little sketch over at the right-hand side, when I had an đť‘Ą-coordinate of zero, it gave me a positive result when I put it into that expression.
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When I have an đť‘Ą-coordinate of nought point three, I plug that into the expression, I get a negative result.
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So somewhere between đť‘Ą is zero and đť‘Ą is nought point three, there must be a value of đť‘Ą which generates a đť‘¦-coordinate of zero.
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So now I know that none of this area over here is any use.
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So what Iâ€™m gonna do now is move the left-hand in a little bit.
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And Iâ€™m gonna go up to nought point two.
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Iâ€™m gonna try nought point two as an đť‘Ą-value and see whether that gives me a good answer or not.
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So plugging nought point two into our expression, Iâ€™ve got two times nought point two squared minus five times nought point two plus one.
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And when I evaluate that, I get an answer of nought point nought eight.
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So an đť‘Ą-coordinate of nought point two generates a đť‘¦-coordinate of nought point nought eight.
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Now thatâ€™s-thatâ€™s above zero, so thatâ€™s too big.
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Now nought point two generates an answer which is pretty close to zero, so weâ€™re quite close to the right answer.
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But unfortunately, thatâ€™s not quite good enough a guess.
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So looking at our graph, weâ€™ve updated it; weâ€™ve zoomed in.
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So when we had an đť‘Ą-value of zero, the đť‘¦-coordinate was one.
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An đť‘Ą value of nought point two, gave us a đť‘¦-coordinate of nought point eight.
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So the value of đť‘Ą which generates a đť‘¦-coordinate of zero canâ€™t be in this region over here.
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Now the value of negative nought point three two was generated when I put in an đť‘Ą-coordinate of nought point three.
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So if we look at this, the value of đť‘Ą which generates a đť‘¦-coordinate of zero must be between zero point two and zero point three.
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The answer goes from being too big when we put in đť‘Ą equals nought point two to being too small when we put in đť‘Ą equals nought point three.
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So, Iâ€™ve got two consecutive answers to one decimal place, which is what they asked for in the question.
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The answerâ€™s somewhere between nought point two and nought point three.
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Weâ€™ve gotta find out which one is it closer to.
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Is it closer to nought point two or is it closer to nought point three?
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So with nought point two generating an answer thatâ€™s too big and nought point three generating an answer thatâ€™s too small, the solution for đť‘Ą must be somewhere between the two.
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So we can update the range there.
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Now itâ€™s very tempting at this stage to say, â€śWell nought point two gives us an answer of nought point nought eight, which is very close to zero.
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Nought point three gives us an answer of negative nought point three two, which is a bit further away from zero.
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So the answer must be nought point two, because it generates an answer closer to the answer weâ€™re looking for.â€ť
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But you canâ€™t really do that.
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Now thereâ€™s some sort of a curve representing the đť‘¦-values of this expression in between nought point two and nought point three.
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Now that curve could be more or less a straight line coming down like this, it could be a curve that goes in this direction and comes across here, or it could be a curve that goes across here and down.
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We donâ€™t really know.
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Now if it was this sort of a curve, actually the answer, the đť‘Ą value that generates the đť‘¦-coordinate of zero would be closer to nought point three than it would be to nought point two.
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If itâ€™s either of these two scenarios, then the đť‘Ą-coordinate here is going to be closer to nought point two than it is to nought point three.
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So given that we donâ€™t know what the curve looks like, we have to be a bit clever about this.
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So what we have to do is look at the đť‘Ą-coordinate thatâ€™s midway between nought point two and nought point three.
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Now if that generates an answer which is bigger than zero, then we know that the change from being too big to too small is happening in this range over here.
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If it generates an answer which is too small, then we know the change from being too big to being too small is happening in this range over here.
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So plugging in the đť‘Ą-value nought point two five to our expression, weâ€™ve got two times nought point two five squared minus five times nought point two five plus one.
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And that equals negative nought point one two five, which is too small.
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So in between nought point two five and nought point three, we suspect all of the results are gonna be too small.
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But the change from being too big to too small happens in this range here, between nought point two and nought point two five.
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So being to the left of nought point two five means that itâ€™s definitely closer to nought point two than it was to nought point three.
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So weâ€™ve narrowed down the range to be in between nought point two and nought point two five, which means our answer, correct to one decimal place, is nought point two.
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Now to get full marks on these sorts of questions, we need to see some of these values, a few of these values, correctly evaluated for different values of đť‘Ą within the range specified in the question.
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We also need to see that youâ€™ve narrowed it down to â€” So if weâ€™re going for one decimal place, weâ€™d need two consecutive answers to one decimal place like nought point two and nought point three, and they need to be correctly evaluated as well.
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And then finally, you need to show that youâ€™ve correctly proved whether weâ€™re to the left or to the right of that midpoint of those two consecutive answers, nought point two and nought point three.
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You wonâ€™t get the credit for getting a correct answer, nought point two, unless youâ€™ve done this extra stage here where weâ€™ve gone to one more decimal place, two decimal places in this case, to prove whether weâ€™re closer to nought point two or nought point three.
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Now, the next part of the question is to refine that answer, correct to two decimal places.
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So the expression weâ€™re trying to solve here is two đť‘Ą squared minus five đť‘Ą plus one equals zero.
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And weâ€™d narrowed it down to đť‘Ą was between nought point two and nought point two five.
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So given that we want two decimal places, weâ€™ve - weâ€™ll change that nought point two to nought point two zero and weâ€™ll do another table.
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Right.
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So we just filled in what we know already.
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The result when đť‘Ą is nought point two gives us an answer of nought point nought eight when we evaluate that expression, and thatâ€™s too big.
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And when we plug in đť‘Ą equals nought point two five, that gives us a value for the expression which is below zero, so itâ€™s too small.
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So just quickly sketching out our graph, we know that somewhere between those two đť‘Ą values, weâ€™re gonna find a coordinate which generates a đť‘¦-coordinate of zero: a value for that expression of zero.
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Now, Iâ€™m gonna try an đť‘Ą-value of nought point two two.
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So youâ€™ve got a range, you could try nought point two one, two two, two three, or two four.
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So weâ€™ll go with nought point two two.
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And plugging that into the expression, weâ€™ve got two times nought point two two squared minus five times nought point two two plus one.
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And when we evaluate that, we get an answer of negative nought point nought nought three two.
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So itâ€™s quite close to zero, but itâ€™s below the đť‘Ą-axis.
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Itâ€™s a negative number, so itâ€™s too small.
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So this means that the đť‘Ą-coordinate has to be somewhere between nought point two zero and nought point two two.
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And that will generate a đť‘¦-coordinate of zero.
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So letâ€™s try nought point two one.
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And two times nought point two one squared minus five times nought point two one plus one gives us a value of nought point nought three eight two.
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So thatâ€™s above zeros, so thatâ€™s too big.
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So this means nought point two one is to the left of our answer, and nought point two two is to the right of our answer.
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So we need to try the value thatâ€™s exactly halfway between those and see what result we get.
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So plugging in the next value of nought point two one five gives us two times nought point two one five squared minus five times nought point two one five plus one, which is nought point nought one seven four five.
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Itâ€™s positive.
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Itâ€™s too big.
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So just re-sketching our little graph there, nought point two one generated a đť‘¦-value which was too big.
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Nought point two one five has also generated a đť‘¦-value which is too big.
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So all of those đť‘Ą-values between nought point two one and nought point two one five are gonna generate đť‘¦-values that are too big.
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So the change from being too big to being too small happens between nought point two one five and nought point two two.
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This means that itâ€™s to the right of nought point two one five and therefore itâ€™s closer to nought point two two than it is to nought point two one.
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So our answer is đť‘Ą is nought point two two correct to two decimal places.
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So letâ€™s just check we got everything we need.
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We wanted the answer correct to two decimal places.
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Weâ€™ve got two consecutive answers, one thatâ€™s too small and one thatâ€™s too big.
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So nought point two one and nought point two two, theyâ€™re consecutive answers to two decimal places.
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We checked the value in the middle, and that was too big.
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And that told us that the change from being too big to being too small happens between nought point two one five and nought point two two, which means that the đť‘Ą-value must be closer to nought point two two than it is to nought point two one.
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Thatâ€™s how we got our answer.
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So thatâ€™s full marks.
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Again, it wouldâ€™ve been tempting to say, â€śWell at nought point two one, the đť‘¦-value generated is nought point nought three eight two.
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But at nought point two two, the đť‘¦-value generated is minus nought point nought nought three two.
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Now that is closer to zero than that is.â€ť
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That argument on its own is not good enough to get you those last two marks for the question.
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You have to test the value in the middle and prove whether itâ€™s gonna be to the left or the right of that value.
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So just before we go, letâ€™s do a slow motion action replay of what weâ€™ve just done on the graph.
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We started off evaluating the expression for đť‘Ą equals zero and đť‘Ą equals one and found out that one was too big and one was too small.
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So the answer must be somewhere between zero and one.
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So we then tried an đť‘Ą-value of nought point three and the-the đť‘¦-value that that generated was too small.
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So at this point, we knew that our đť‘Ą-coordinate must be between zero and zero point three.
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We then tried an đť‘Ą-coordinate of nought point two.
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That generated an answer that was too big.
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So now, we knew the answer was between zero point two and zero point three.
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So we checked out the mid đť‘Ą-value between nought point two and nought point three, thatâ€™s at nought point two five, and found that that gave us an answer which was too small.
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And this told us that the answer must be between nought point two and nought point two five because thatâ€™s where the answer goes from being too big to being too small.
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We wanted our answer to one decimal place, so itâ€™s either gonna nought point two or nought point three.
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And by doing this midpoint value and working our way to the left of that, we know that the final answer must be nought point two.