Introductory Engineering Mathematics by David Reeping & Kenneth Reid

Author:David Reeping & Kenneth Reid
Language: eng
Format: epub
Publisher: Momentum Press
Published: 2017-04-17T04:00:00+00:00

Do not be tempted to simplify this any further; we just brought out the number we needed to factor! Notice that the left-hand side, x2 + 8x + 16, is a perfect square?

Now we can easily solve for x:

Therefore, either x = 0.6904 or x = –8.6904 will give us a value for f (x) = 0—this means the roots of this function are 0.6904 and –8.6904.

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3.3.1 LOCATING ROOTS WITHIN A TOLERANCE, OR “GETTING CLOSE ENOUGH”

We did not need to use “completing the square” in order to solve the problem in Example 3.10, but it does provide insight into the ways in which we can play around with the expression in order to arrive at the solution. The most standard formula most of us know by heart is the quadratic formula, which enables us to find the exact roots to second-degree polynomials (quadratic equations). We know a quadratic equation, ax2 + bx + c = 0 where a, b, and c are constants, has the following solutions:

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