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Use a trial-and-improvement method to solve the equation 𝑥 cubed plus two 𝑥 squared minus 25 equal zero to one decimal place, knowing that the equation has a solution between two and three.
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In this question, we’re asked to solve the equation.
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This means that we need to find the value of 𝑥 for which 𝑥 to the third power plus two 𝑥 squared minus 25 would equal zero.
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We can solve equations in a number of ways, for example, by factoring or graphically.
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However, here we’re asked to solve using trial and improvement.
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This means that we’re going to try different values of 𝑥 to see if we can get our equation close to zero.
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We can start by using any values of 𝑥.
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For example, 𝑥 is negative 10 or 𝑥 is 100.
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But we’re actually given a starting point that 𝑥 is between two and three.
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So we’re going to start with those values.
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The important thing with the trial-and-improvement method is finding a neat way to present our workings.
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And we can do this using a table.
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We start with a column for the value of 𝑥 that we’re going to be using.
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For example, the first trial will be 𝑥 equals two.
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We need a column for calculating 𝑥 to the third power plus two 𝑥 squared minus 25 for our value of 𝑥.
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And it may be helpful to add three columns in between those two to break down the equation into smaller pieces.
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And finally, we add a column to record whether the value of our equation is too big or too small in comparison to zero.
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So let’s then start with our first trial of 𝑥 equals two, since we were told that the equation has a solution between two and three.
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We substitute 𝑥 equals two into our equation.
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So that means that every value of 𝑥 here will be replaced with the value two.
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We can recall that two cubed is the same as two times two times two, which is eight.
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When we have two times two squared, this is the same as two times two times two, which is also equal to eight.
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And next, the value of negative 25 will always stay the same.
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So to work out the value of our equation, we take the three pieces of information on 𝑥 to the third power, two 𝑥 squared, and negative 25, which is eight plus eight take away 25.
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So the value of our equation will be negative nine.
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We could also have put the whole equation with 𝑥 equals two into our calculator, which would’ve given us the same value of negative nine.
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We now need to check if our value of negative nine is bigger or smaller than zero.
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And since it’s less than zero, we know that our value of 𝑥 must be too small.
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So for our next trial, we’re going to use the value of 𝑥 equals three.
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So then three to the third power will be three times three times three, which is 27.
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Next, two times three squared will be the same as two times three times three, giving us 18.
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And our negative 25 stays the same.
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Putting it together then, we have 27 plus 18 minus 25, which will give us an answer of 20.
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Next, how is our value of 20 compared with the value of zero?
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Well, it’s too big.
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And note that this does also confirm that our equation does have a solution between two and three.
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So now let’s try a value of 𝑥 that’s smaller than three.
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We could use the value of 𝑥 equals 2.5 since that’s nicely in between two and three.
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We can start by calculating 2.5 to the third power, which we can evaluate as 15.625.
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Next, two times 2.5 squared will give us 12.5.
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And our negative 25 stays the same.
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Putting it together, we have 15.625 plus 12.5 minus 25, which is equal to 3.125.
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And then since 3.125 is bigger than zero, our value of 𝑥 must be too big.
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This means, for our next value of 𝑥, we should choose a number that’s less than 2.5.
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Let’s try 𝑥 equals 2.4.
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At this point, we can add a handy calculator tip, particularly if you’ve been working out the whole of the equation in one go.
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If your calculator has a replay function, this means that you can change the values that were last entered to a different value.
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In this case, you could change all the values that said 2.5 into 2.4.
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For each individual term, when 𝑥 equals 2.4, we would have 13.824, 11.52, and negative 25.
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The value of our equation will be 0.344.
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And since this is larger than zero, then our 𝑥-value is too big.
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We can notice, however, that our equation is quite close to zero.
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So our next smaller value of 𝑥 shouldn’t be too much smaller than 2.4.
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So let’s choose 𝑥 equals 2.3.
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Carrying out our calculations will give us 𝑥 to the third power plus two 𝑥 squared minus 25 as negative 2.253.
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So this means that the value of 𝑥 is too small.
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However, we can notice that we have now narrowed it down to one decimal place.
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Since we have 𝑥 equals 2.3 is too small and 𝑥 equals 2.4 is too big, then our value of 𝑥 must be somewhere in between these values.
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We were asked for an answer to one decimal place.
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So how do we choose between 2.3 and 2.4?
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And the answer is that we choose the value in between these two to check.
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And since that value is 2.35, then we carry out one more trial with 𝑥 equals 2.35 and check the result.
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And the value of our equation with 𝑥 equals 2.35 will be negative 0.977125.
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The important thing here is that this value is too small.
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If we visualise this on a number line, we can now see that 𝑥 must be between 2.35 and 2.4.
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If we took any of the values in this range and write them down to one decimal place, we would get 2.4.
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This means that our solution to one decimal place is 2.4.