CHAPTER XXXIV. THEORY OF QUADRATIC EQUATIONS. MISCELLANEOUS THEOREMS. 300. In Chapter xxvi. it was shown that after suitable reduction every quadratic equation may be written in the form ax2+bx+c=0 ..... .(1), and that the solution of the equation is -6+ V62-4ac :(2). 2a We shall now prove some important propositions connected with the roots and coefficients of all equations of which (1) is the type. NUMBER OF THE Roots. 301. A quadratic equation cannot have more than two roots. For, if possible, let the equation ax2+bx+c=0 have three different roots a, b, y [Art. 218]. Then since each of these values must satisfy the equation, we have aa'+ba+c=0..... ..(1), aß? + bB+c=0..... ..(2), ay+by+c=0. .. (3). From (1) and (2), by subtraction, a (a-)+(a-B)=0; divide out by a-ß which, by hypothesis, is not zero; then a(a+)+b=0. Similarly from (2) and (3) a (B+y)+b=0; .: by subtraction u(a-)=0; which is impossible, since, by hypothesis, a is not zero, and a is not equal to y. Hence there cannot be three different roots. 302. The terms unreal,' imaginary,' and `impossible' are all used in the same sense: namely, to denote expressions which involve the square root of a negative quantity, such as ✓-1, -3, -a. It is important that the student should clearly distinguish between the terms real and rational, imaginary and irrational, Thus V25 or 5, 31, - are rational and real; v7 is irrational but real; while V-7 is irrational and also imaginary. a CHARACTER OF THE Roots. 303. In Art. 300 denote the two roots in (2) by a and B, - 6+ V62 - 4ac -O-V62 - 4ac B=2a 2a then we have the following results : (1) If b2-4ac, the quantity under the radical, is positive, the roots are real and unequal. (2) If b2-4ac is zero, the roots are real and equal, each 6 reducing in this case to 2a (3) If 62–4ac is negative, the roots are imaginary and unequal. (4) If b2 – 4ac is a perfect square, the roots are rational and unequal. By applying these tests the nature of the roots of any quadratic may be determined without solving the equation. Example 1. Show that the equation 2x2 - 6x+7=0 cannot be satisfied by any real values of x. Here a=2, b=-6, 6=7; so that 62—4ac=(-6)2–4.2.7=-20. Therefore the roots are imaginary. Example 2. For what value of k will the equation 3x2—6x+k =0 have equal roots ? The condition for equal roots gives (-6)2–4.3. k=1, whence k=3. Example 3. Show that the roots of the equation x2 — 2px +p2-92+2qr-y2=0 are rational. The roots will be rational provided (-2p)2–4(22—(+2qr -- y2) is a perfect square. But this expression reduces to 4(72—29r+r2), or 4(q-r)?. Hence the roots are rational. RELATIONS OF RootS AND COEFFICIENTS. a aß= -6+ 162 - 4ac -b-V62-4ac 304. Since a= 2a 2a we have by addition -6+ V62-4ac-b-V62 — 4ac b a+ß .(1); 2a and by multiplication we have (-6+ Vb2 – 4ac)(–6– Vb2 — 4ac) 4a? (-6)2–(62 – 4ac) 4a2 4ac (2). 4a2 b By writing the equation in the form x2 + x+-= -0, these results may also be expressed as follows: In a quadratic equation where the coefficient of the first term is unity, (i) the sum of the roots is equal to the coefficient of x with its sign changed; (ii) the product of the roots is equal to the third term. Note. In any equation the term which does not contain the unknown quantity is frequently called the absolute term. с a с a с a a b the equation x2+ -*+- may be written 22-(a+b)x+aß=0... (1). Hence any quadratic may also be expressed in the form 22 – (sum of roots) x + product of roots=0. .... (2). Again, from (1) we have (x-a)(x-3)=0... (3). We may now easily form an equation with given roots. or Example 1. Form the equation whose roots are 3 and -2. 22 -- X – 6=0. 7 The equation is =0); ori (--) (+) or that is, (7x – 3) (5x+4)=0, 35x2 + 13x – 12=0. When the roots are irrational it is easier to use the following method. Example 3. Form the equation whose roots are 2+13 and 2 -3. We have sum of roots=4, product of roots=1; .. the equation is z? – 4x+1=0, by using formula (2) of the present article. 306. The results of Art. 304 are most important, and they are generally sufficient to solve problems connected with the roots of quadratics. In such questions the roots should never be considered singly, but use should be made of the relations obtained by writing down the sum of the roots, and their product, in terms of the coefficients of the equation. Example 1. If a and ß are the roots of x2 – px +q=0, find the value of (1) a2 +ß?, (2) a+B3. We have a+B=P, aß=q. =p2 – 29. Again, a3 +B3 = (a +B) (a+ B2 – aß) =p {(a+) - 3aB} =p (p2 – 39). Example 2. If a, ß are the roots of the equation lx2 + mx +n=0, B find the equation whose roots are B' B a” +32 We have sum of roots = a3 + a 81 81 product of roots or and aß=ī 1 n n .. by Art. 305 the required equation is la+ x2 x+1=0, αβ aßx2 – (a? +B2) x + aß=0. As in the last example a2 +82 ma --2nl 22 m2 – 2nl the equation is 12 c+7=0, nlx? – (m2 – 2nl) x +nl=0. Example 3. Find the condition that the roots of the equation ax2 +bx+c=0 should be (1) equal in magnitude and opposite in sign, (2) reciprocals. The roots will be equal in magnitu and opposite in sign if their b sum is zero; therefore =0, or b=0. or a Again, the roots will be reciprocals when their product is unity; therefore -1, or c=a. с a Example 4. Find the relation which must subsist between the coefficients of the equation px? +qx+r=0, when one root is three times the other. We have a+p= q P but since a=3B, we obtain by substitution 1 T or qa 16p23p' 3qo=16pr, which is the required condition. 307. The following example illustrates a useful application of the results proved in Art. 303. |