a You can refer to a specific cell of the spreadsheet by naming the letter of the column and the number of the row that it belongs to.
B
b Recall that a solution of an equation is the value for which the equation holds true.
C
c Look at the completed spreadsheet from Part B. Does anything characteristic happen at the values near the solution?
A
a =6*(A2)^2-24
B
b See solution.
C
c See solution.
Practice makes perfect
a In the exercise we are shown the spreadsheet shown below.
We can use the spreadsheet's tools to evaluate the left-hand side of the equation. To do this, we can refer to a specific cell by naming the letter of the column and the number of the row that it belongs to. Then, to evaluate the left-hand side of the given equation at x=-3, we can use the expression =6*(A2)^2-24.
After pressing enter the result should appear.
b In Part A we already found a formula to evaluate the expression for the value x = -3 in the cell B2. If we select that cell, a small square will appear in the corner and the formula will be indicated in the row labeled fx.
If we drag the small square from the cell B2 to B8, we indicate that we want to use the same formula for the rest of the x-values and the expression's values are shown immediately in the corresponding cells.
Notice that for x=2 and x=-2 the left-hand side of the equation is 0, just as the right-hand side. Therefore, since these values make the equation hold true, they are the solutions to the equation.
c From Part B we know that the solutions are the values that make the left-hand side equal to 0. Furthermore, note that before and after each 0 the expression's value changes sign.
Therefore, if the solution is not an integer we can still know between which two integers the solutions are by identifying any change in the signs.
Let's consider the example equation, 16x^2-121 =0.
We can see that at no integer value does the expression become zero. However, the expression's values changes sign two times. These changes happen between - 3 and - 2, and between 2 and 3.
Hence, we know that the solutions are in the intervals -3 < x < -2 and 2
Now, we can ensure that the solutions lie in the intervals -3 < x < -2.5 and 2.5 < x < 3. Our new approximations would be x = -2.75 and x = 2.75. If we wanted to make our estimation even more precise, we could repeat the same process.
