Hence, a+√b=a+ √a2—a2+b=a+√b. The second formula can be verified in the same manner. 157. REMARK. As the accuracy of the formulas (6) and (7) is proved, whatever may be the quantity c, or √a2-b, it follows, that when this quantity is not a perfect square, we may still replace the expressions Va+√b and √a−√b, by the second members of the equalities (6) and (7); but then we would not simplify the expression, since the quantities p and q would be of the same form as the proposed expression. We would not, therefore, in general, use this transformation, unless a2-b is a perfect square. EXAMPLES 158. Take the numerical expression 94+42 √5, which reduces to 94+ √8820. We have whence a=94, b=8820, c= √a2―b= √/8836—8820—4, a rational quantity; therefore the formula (6) is applicable to this In fact, (7+3√5)2=49+45+42√5=94+42 √5. Again, take the expression we have whence and np+2m2-2m √np+m2; a=np+2m2, b=4m2(np+m2), a2-b=n2p2, cor Va2-b-np ; therefore the formula (7) is applicable. It gives for the required In fact, +(√np+m2-m). np+m3―m)2=np+2m2-2m √np+m2. For another example, take the expression and reduce it to its simplest terms. By applying the preceding formulas, we find 16+30 √−1=5+3 √-1, √16-30 √-1-5-3 √—1. Hence, 16+30 √−1+▼ 16-30 √-1=10. This last example shows, better than any of the others, the utility of the general problem; because it proves that imaginary expressions combined together, may produce real, and even rational results. 28+10 √3 =5+ √3 ; √1+4 √ −3 =2+ √ −3, bc+2b √ bc-b2 + √ bc−2b √ bc—b2 =±2b; √ ab+4c2 d2+2 √ 4abc2_abd2= √ab+ √4c2_d2. - Examples of Equations of the Second Degree, which either involve Radicals, or two unknown quantities. |