Video: Pack 4 β€’ Paper 3 β€’ Question 15

Pack 4 β€’ Paper 3 β€’ Question 15

07:24

Video Transcript

Show that there is a solution to the equation π‘₯ cubed plus nine π‘₯ minus four equals zero between π‘₯ equals zero and π‘₯ equals one.

To answer this question, we need to evaluate the function π‘₯ cubed plus nine π‘₯ minus four at both π‘₯ equals zero and π‘₯ equals one. When π‘₯ is equal to zero, we have zero cubed plus nine multiplied by zero minus four, which is equal to negative four. When π‘₯ is equal to one, we have one cubed plus nine multiplied by one minus four. This gives one plus nine minus four which is six.

The key point here is that the function π‘₯ cubed plus nine π‘₯ minus four takes a negative value when π‘₯ is equal to zero and a positive value when π‘₯ is equal to one. As π‘₯ cubed plus nine π‘₯ minus four is a continuous function, which means there are no gaps, this means that it must be equal to zero somewhere between π‘₯ equals zero and π‘₯ equals one as its value has crossed over from negative to positive between these values. This method of showing us that the equation has a solution between two values is sometimes called the change of sign method.

Rearrange the equation π‘₯ cubed plus nine π‘₯ minus four equals zero into the form π‘₯ equals π‘Ž over π‘₯ squared plus 𝑏.

We’ve been asked to rearrange the equation from a form where the highest power of π‘₯ is a three to a form where the highest power is a two. This suggests that we should factorize by π‘₯. However, the third term negative four doesn’t have a factor of π‘₯. So we can only factorize from the first two terms. This gives π‘₯ multiplied by π‘₯ squared plus nine minus four is equal to zero.

Now, this is starting to look a little bit more like what we’re hoping to rearrange to as we have a bracket π‘₯ squared plus nine, which looks like the denominator of the fraction. Next, we can add four to both sides of the equation, giving π‘₯ multiplied by π‘₯ squared plus nine is equal to four.

The final step is to divide both sides of the equation by that bracket β€” π‘₯ squared plus nine β€” giving π‘₯ is equal to four over π‘₯ squared plus nine. Now, we aren’t explicitly asked to state the values of π‘Ž and 𝑏. But comparing our equation to the form we were asked for, we can see that the value of π‘Ž is four and the value of 𝑏 is nine.

Use the iterative formula π‘₯ 𝑛 plus one is equal to π‘Ž over π‘₯ 𝑛 squared plus 𝑏 with the values of π‘Ž and 𝑏 found above to estimate a solution to π‘₯ cubed plus nine π‘₯ minus four is equal to zero. Start with π‘₯ zero equals zero and use the iterative formula three times. Give your answer to three decimal places.

First, let’s recall our answer to the previous part of the question, which was that π‘₯ is equal to four over π‘₯ squared plus nine. This means that the value of π‘Ž in our iterative formula is four and the value of 𝑏 is nine.

Now, let’s recall what an iterative formula means. It’s used to estimate the solution to an equation by progressively improving the estimate. It tells us that to find the next value of π‘₯, π‘₯ 𝑛 plus one, we just substitute the current value of π‘₯, π‘₯ 𝑛, into this formula. As this equation is a rearrangement of the equation π‘₯ cubed plus nine π‘₯ minus four equals zero, the values of π‘₯ generated by this iterative formula will give an estimate of a solution to the equation π‘₯ cubed plus nine π‘₯ minus four equals zero.

We’re told that the starting value for our estimate π‘₯ zero is equal to zero. To find the value of π‘₯ one, we substitute the value of π‘₯ zero into the iterative formula, giving four over zero squared plus nine. This is equal to four-ninths or as a decimal it’s equal to 0.4 recurring.

To find the next value of π‘₯, π‘₯ two, we substitute our current value of π‘₯ into the formula, giving four over four-ninths squared plus nine. As a decimal, this gives 0.434899 and the three dots indicate that the decimal continues. To perform this accurately on your calculator, particularly, if you’re using the iterative formula several times, you can use the previous answer button.

If you use the fraction and brackets keys on your calculator, you can input four over answer squared plus nine. And your calculator will then take the value on screen, square it, add nine, and then divide four by this. This means that you can just keep pressing enter and your calculator will keep giving you the next value in this iterative sequence.

We’re told to use the iterative formula three times. So we need to perform one more iteration. The calculation for π‘₯ three is four over 0.4348 squared plus nine. But in reality, you can just press equals on your calculator if you’ve used the previous answer button. This gives a decimal of 0.43529656 and the decimal continues.

We’re asked to give our answer to three decimal places. And as the deciding number β€” so that’s the fourth decimal place β€” is a two, we round down. Our estimate of a solution to the equation π‘₯ cubed plus nine π‘₯ minus four equals zero to three decimal places is 0.435.

Substitute your value for π‘₯ three into π‘₯ cubed plus nine π‘₯ minus four. Comment on how accurate your estimate is to the solution of π‘₯ cubed plus nine π‘₯ minus four is equal to zero.

Our estimate of π‘₯ three is 0.435. So substituting this into the given expression, we have 0.435 cubed plus nine multiplied by 0.435 minus four. Evaluating this on a calculator gives an answer in standard form: negative 2.687125 multiplied by 10 to the power of negative three.

Multiplying a number by 10 to a negative power means that the number is getting smaller. In this case, it means that we are dividing by 10 to the power of three or 1000. Dividing negative 2.687125 by 1000 gives the number negative 0.002687125, which has its decimal point three places left of where it was in the original number.

We’re asked to comment on how close our estimate is to the solution of π‘₯ cubed plus nine π‘₯ minus four is equal to zero. Since the value we get when we substitute our estimate into this expression is very close to zero, this means that our value of π‘₯ three is close to the solution.

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