Squaring this, x2—1+2x3—x2—x+1+x-1=x3. By transposing and changing the signs, x3—x2—x+1—2√✓✅x3—x2—x+1=—1. Herex3-x2-x+1, may be considered as the un x3-x-x+1-1-0, or x3-x-x+1=1. 6. Given 3x-2x+3=11, to find x. 7. Given x4n ——— -2x3n+x2=132, to find the value of x. This equation may be changed to x4n-2x3n+x2n—x2n+xn=132. Which is (1)2 — (x21‚—x1)—152. 8. Given x2+12x=64. Required the value of x. 9. Given 7x+5x+4=82, to find x. 1 10. Given —~——~+7=8, to find x. 11. Given 2+20x3-224, to find x. 13. Given 6x+ Ans. x=4. Ans. x=3. Ans. 2-11 Ans. x=2. Ans. x=6. =44, to find x. Ans. x=7, or 5.* *This example falls under the ambiguous case, which is ex plained in the next note. 62. The foregoing solutions have been effected by executing the whole process in each case; but it is more convenient in practice, as well as more elegant, to solve all such equations by a few general formula. In order to which, it is to be observed, that all adfected quadratic equations are reducible to one of the following forms. These equations, when the squares are completed, be come, 1. x2+2ax+a2=a2+b 2. x2-2ax+a2=a2+ 3. x2-2ax+a°—a3—b. Hence, by evolution and transposition, we have 1. x=—a+√a2+b. 2. x=a+√a2+b. 3. x=a±✓a2—b.* Adfected quadratic equations are sometimes expressed by one of the following forms. Multiplying these equations severally by 4a, and adding b2 to each member, they become, 1. 4a2x2+4abx+b2=b2+4ac. 2. 4a2x2-4abx+b2=b2+4ac. 3. 4a2x2-4abx+b2=b2-4ac. The roots being extracted, the values of x are easily found. From these formula, any adfected quadratic equation may be resolved, by assuming for a and b, in the equations (A,) and for a, b, and c, in the equations (B,) such values as will render the general equation identical with the particular one under consideration, and substituting the values thus assumed in the final equation. Simple quadratic equations are evidently solvible by the same formulæ, if we assume the co-efficient in the second term =0. * The value of x, in the third case, is ambiguous; because, ✔x2-2ax+a2=x-a, or a-x, and 4a2x2-4abx+b2=2ax -b, or b―2ax; and there is nothing in the general equation, to show whether x is greater or less than a in the former case, or b than in the latter. The ambiguity does not exist in the second 2a case, because a+b must be greater than a, and b3+4ac greater than b, and therefore, cannot be equal respectively to a-x, and b-2ax, unless x be negative. When the sign of x is doubtful, the first and second cases are also ambiguous. Comparing this equation with the 2d form, (A,) we have Or comparing the given equation with the 2d form (B.) 18. Given 10+x-10+x=2, to find the value of x. Considering 10+ as the quantity sought, and comparing the equation with the 2d form, (A,) we have Hence by involution, 10+x-4✅/10+x+4=√10+x. Hence, 14+x=5/10+x. By involution, 196+28x+x3-250+25x. Hence, x2+Sx=54. Comparing this with the 1st form, (A,) |