If we represent the coefficient of x by 2p, and the known term by q, we have a2+2px=9, an equation containing but three terms; and we see, from the above examples, that every complete equation of the second degree may be reduced to this form. 123. We wish now to show that there are four forms under which this equation will be expressed, each depending on the signs of 2p and q. 1st. Let us for the sake of illustration, make 2p+4, and q=+5: we shall then have x2+4x=5. 2nd. Let us now suppose 2p=-4, and q=+5: we shall then have x2-4x=5. QUEST.-123. Under how many forms may every equation of the second degree be expressed? On what will these forms depend? What are the signs of the coefficient of x and the known term, in the first form? What in the second? What in the third? What in the fourth! Repeat the four forms. We therefore conclude that every complete equation of the second degree may be reduced to one of these forms: x2+2px=+q, 1st form. x2-2px=+7, 2nd form. x2+2px=-4, 3rd form. x2-2px-q, 4th form. 124. REMARK.-If, in reducing an equation to either of these forms, the second power of the unknown quantity should have a negative sign, it must be rendered positive by changing the sign of every term of the equation. 125. We are next to show the manner in which the value of the unknown We have seen (Art. 38), that dions, we ay be found. +75=x2 (x+P^ virx_px+p2 ; and comparing this square with the first and third forms, we see that the first member in each contains two terms of the square of a binomial, viz: the square of the first term plus twice the product of the 2nd term by the first. If, then, we take half the coefficient of x, viz: p, and square it, and add to both members, the equations take the form x2+2px+p2=q +p2, x2+2px+p2=−q2+p2, in which the first members are perfect squares. This is QUEST.-124. If in reducing an equation to either of these forms the coefficient of x is negative, what do you do?-125. What is the square of a binomial equal to ? What does the first member in each form contain? How do you render the first member a perfect square? What is this called? called completing the square. Then, by extracting the square root of both members of the equation, we have 126. If we compare the second and fourth forms with Then, by extracting the square root of both members, we have QUEST.-126. In the second form, how do you make the first mem ber a perfect square? 127. Hence, for the resolution of every equation of the second degree, we have the following RULE. I. Reduce the equation to one of the known forms. II. Take half the coefficient of the second term, square it, and add the result to both members of the equation. III. Then extract the square root of both members of the equation; after which, transpose the known term to the second member. REMARK.-The square root of the first member is always equal to the square root of the first term, plus or minus half the coefficient of x. If we first divide by the coefficient 2, we obtain QUEST.-127. Give the general rule for resolving an equation of the second degree. What is the first step? What the second? What the third? What is the square root of the first member always eqnal to? That is, in this form the smaller root is positive, and the larger negative. Verification. If we take the positive value, viz: x=+4, and if we take the negative value of x, viz: x=-8, the equation x2+4x=32 gives (−8)2+4(-8)=64-32=32. . From which we see that either of the values of x, viz: x=+4 or x=-8, will s 4x=5. 2. What are the values of equation. 3x2+12x-19=-x2-12x+89 ? By transposing the terms, we have 3x2+x2+12x+12x=89+19: |