Elementary algebra for schools

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1885 Excerpt: ...6x+ (3)2--25, or #2-6#=16. Hence, by retracing our steps, we learn that the equation x-6x = 16 can be solved by first adding (3)2 or 9 to each side, and then extracting the square root; and the reason why we add 9 to each side is that this quantity added to the left side makes it a perfect square. 194. We have shewn that the square may readily be completed when the coefficient of x2 is unity. All cases may be reduced to this by dividing the equation throughout by the coefficient of xl. Example 1. Solve 32-3a?=10x. Transposing, 3x2 + 10x = 32. Divide throughout by 3, so as to make the coefficient of a? unity. Thus 3 + = f, 195. We see then that the following are the steps required for solving an adfected quadratic equation: (1) If necessary, simplify the equation so that the terms in x2 and x are on one side of the equation, and the term without x on the other... (2) Make the coefficient of x2 unity and positive by dividing throughout by the coefficient of x2. (3) Add to each side of the equation the square of half the coefficient of x. (4) Take the square root of each side. (5) Solve the resulting simple equations. 196. In the examples which follow some preliminary reduction and simplification may be necessary. Example 1. Solve--5=_-V-2 „....." 3x-2 3x-8 Simplifying, 197. In all the instances considered hitherto the quadratic equations have had two roots. Sometimes, however, there is only one solution. Thus if x2-2x +1 = 0, then (x-1 )2 = 0, whence 37=1 is the only solution. Nevertheless, in this and similar cases we find it convenient to say that the quadratic has two equal roots. 1. 4. 7. 10. 12. 14. 16. 18. 20. 22. 24. 25. 28. 30. 32. 34. 198. From the preceding examples it appears that after suitable reduction and transposition every quadrati...