Elements of Algebra

CHAP. VI. OF ALGEBRA. 225 as before. It only remains, therefore, to illustrate this rule by some examples. 646. Question 1. There are two numbers; the one exceeds the other by 6, and their product is 91: what are those numbers? If the less be $x$, the other will be $x + 6$, and their product $x^2 + 6x = 91$. Subtracting $6x$, there remains $x^2 = 91 - 6x$, and the rule gives $x = -3 \pm \sqrt{(9 + 91)} = -3 \pm 10$; so that $x = 7$, or $x = -13$. The question therefore admits of two solutions; By the one, the less number $x = 7$, and the greater $x + 6 = 13$. By the other, the less number $x = -13$, and the greater $x + 6 = -7$. 647. Question 2. To find a number such, that if 9 be taken from its square, the remainder may be a number, as much greater than 100, as the number itself is less than 23. Let the number sought be $x$. We know that $x^2 - 9$ exceeds 100 by $x^2 - 109$: and since $x$ is less than 23 by $23 - x$, we have this equation $x^2 - 109 = 23 - x$. Therefore $x^2 = -x + 132$; and, by the rule, $x = -\frac{1}{2} \pm \sqrt{(\frac{1}{4} + 132)} = -\frac{1}{2} \pm \sqrt{(\frac{529}{4})} = -\frac{1}{2} \pm \frac{23}{2}$. So that $x = 11$, or $x = -12$. Hence, when only a positive number is required, that number will be 11, the square of which minus 9 is 112, and consequently greater than 100 by 12, in the same manner as 11 is less than 23 by 12. 648. Question 3. To find a number such, that if we multiply its half by its third, and to the product add half the number required, the result will be 30. Supposing the number to be $x$, its half, multiplied by its third, will give $\frac{1}{6}x^2$; so that $\frac{1}{6}x^2 + \frac{1}{2}x = 30$; and multiplying by 6, we have $x^2 + 3x = 180$, or $x^2 = -3x + 180$; which gives $x = -\frac{3}{2} \pm \sqrt{(\frac{9}{4} + 180)} = -\frac{3}{2} \pm \frac{27}{2}$. Consequently, either $x = 12$, or $x = -15$. 649. Question 4. To find two numbers, the one being double the other, and such, that if we add their sum to their product, we may obtain 90. Let one of the numbers be $x$, then the other will be $2x$; their product also will be $2x^2$, and if we add to this $3x$, or their sum, the new sum ought to make 90. So that $2x^2 + 3x = 90$; or $2x^2 = 90 - 3x$; whence $x^2 = -\frac{3}{2}x + 45$, and thus we obtain Q