By J. Parry Lewis (auth.)

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**Sample text**

2. Also solve the following by this method, noting that in this case the figures involved are rather more awkward, and lead to solutions which could not be found by any attempt to factorise: (i) 3x2 +4x-5=0 (iii) 2x2 =3x+7 (ii) x 2 -4x=8 (iv) 2x2 -8x+3=0 20 ALGEBRA Before leaving the subject of quadratic equations there are two points for us to notice. The first is that if we examine the solutions to the equations we have solved here, and to those in the excercises, we will find that in every case the sum of the two solutions is equal to - bfa and their product is equal to cfa.

It will be noticed that we were wise not to proceed further with our attempt to locate the root between - 3 and -4 since this would have involved the labour of cubing - 3·44 (without any guarantee that it would give us the answer). Often it is a good practice in such cases to locate the root approximately, as we did, and then to use some other method, such as Newton's method (given in Appendix 5) to locate it more accurately, and with less trouble than is involved in the repeated application of the Remainder Theorem.

E. either x=1+t=! or x=1-t=t (iii) x+i= J~: :. J:~ = + 1·302 or -1·302 (which is written ± 1·302) . X= -f± 1·302 and so 1. 6 using the method just described. The third method is derived as follows. Consider ax2 +bx+c=0 where a, b and c are any numbers, positive, negative or zero, provided that a is not zero. Let us solve this, finding x in terms of a, b and c, by the method of completing the square. Jb2 -4ac 2a We can now use this result, which should be memorised, to solve a quadratic equation.