# Question Video: Using Trial and Improvement to Find the Solutions to an Equation Mathematics

William is trying to find the solution to the equation 𝑥³ + 8𝑥 = 100. He decides to use trial and improvement and has so far completed the table. Work out the three missing outputs from his table, giving your answer to two decimal places. Hence, work out the solution to the equation to two decimal places.

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### Video Transcript

William is trying to find the solution to the equation 𝑥 cubed plus eight 𝑥 equals 100. He decides to use trial and improvement and has so far completed the table. Work out the three missing outputs from his table, giving your answer to two decimal places. Hence, work out the solution to the equation to two decimal places.

We notice that for trial and improvement, this means that we try different values of 𝑥 to see if we can get 𝑥 cubed plus eight 𝑥 to equal 100. Here, William has started by trying 𝑥 equals four. He substituted this value of 𝑥 into the expression 𝑥 cubed plus eight 𝑥 to give four cubed plus eight times four, giving 64 plus 32 and writing the result of 96 in his table. Comparing this value of 96 with 100, he would’ve realized that his value was two small. He then chose the larger value of 𝑥 equals five to try. Substituting 𝑥 equals five into the expression 𝑥 cubed plus eight 𝑥 would have given the result of 165.

Comparing his found result of 165 with the value of 100, William would have noticed that this value was too large, meaning that his next trial of 𝑥 should be smaller than five but, of course, larger than four. His next trial when 𝑥 equals 4.1 gave the result of 𝑥 cubed plus eight 𝑥 as 101.72. To find the result in the table when 𝑥 equals 4.09, we substitute this value into the expression 𝑥 cubed plus eight 𝑥, which gives us 4.09 cubed plus eight times 4.09. We can then use our calculator to evaluate the result, giving us 101.137929.

We’re asked for the outputs in the table to two decimal places. So, we check if the third decimal digit is five or more. And since it is, then our answer will round up to 101.14. And so, our first missing value in the table is 101.14. As this value is also too big, then we can check a lower value of 4.07. We substitute the value of 4.07 into the expression, giving us 4.07 cubed plus eight times 4.07. Again, using our calculator then, we can evaluate the result as 99.979143.

Rounding our answer to two decimal places will give us the value of 99.98. As this result is too small, then William has tried a slightly larger value of 𝑥, 4.075. Substituting this value in will give us the result of 100.2679219, which is 100.27 to two decimal places. And this value is also too big since it’s larger than 100.

We have now found the three missing outputs for the first question. So, let’s answer the second question, which is finding the solution to the equation. If we were to represent our values of 𝑥 on a number line, then as our value of 4.09 was too big and the value for 4.07 was too small, then the solution must lie between 4.07 and 4.09. In fact, we can make this interval smaller since we also know that 4.075 was too big too. Therefore, our answer is in the range 4.07 to 4.075.

This line makes giving our answer to two decimal places quite simple since we know that every value in this range will round to 4.07. So, our answer would be 𝑥 equals 4.07. Therefore, the answer for the first question, the three missing outputs from the table, is 101.14, 99.98, and 100.27. And the answer for the second question, to work out the solution of the equation, would be 𝑥 equals 4.07.