Adv Maths Question 36

I need correct answer otherwise compulsory I give multiple Dislikes

Question 36 (i) The equation x 2 + bx + c = 0 has the property that if k is a root, then k 1 is a root. Show that either c = 1, or c = 1 and b = 0. (ii) Instead of the above property, the equation x 2 + bx+ c = 0 has the property that if m is a root, then 1 m is a root. Determine carefully the restrictions that this property places on b and c. (iii) The equation x 2 + bx + c = 0 has both the properties described in (i) and (ii) above. Show that b = 1 and c = 1. (iv) The equation x 3 + px2 + qx + r = 0 has both the properties described in (i) and (ii) above. Find the possible values of p, q and r. Discussion This question was difficult to word, because of the possibility that k 1 and k need not be distinct. (For example, one should not write k 1 is also a root in line 1, since this might seem to imply that k is different from k 1 .) Note that the roots are not restricted to being real. You will need to marshal your thoughts carefully for all parts of the problem. In part (i) you have to make sure that you have thought of all the possibilities, remembering that k could be, for example, +1, in which case the given property relates the root to itself rather than to the other root. It is not a bad idea to start with Let the roots be and so as not to keep referring to the other root. No extra work is required for part (iii): you only have to compare the results of parts (i) and (ii). Part (iv) needs careful organisation. One plan is to consider separately the case when the equation has repeated roots. Having dealt with this (using the previous part), you can concentrate on the equations with three different roots. You may have noticed that the transformations which preserve the set of roots in parts (iii) and (iv) (i.e. z z 1 and z 1 z) form a group, the group multiplication being composition of the transformations (perform one transformation, then the next). Call these two transformations f and h respectively. Then we have the structure f2 = h2 = 1 and f hf = hf h (verify this by choosing z = 3, say, or prove it for general z), so the distinct elements are 1, f, h, f h, hf and f hf. The group has order 6 and is isomorphic to the group of permutations of three objects and to the symmetry group (rotations and reflections) of a triangle. You can think of the group acting on points in the Argand diagram, where it will generally map a set of six points into itself; for example, if we start with the point z = 3, we get the set consisting of 3, f(3) = 1 3 , h(3) = 2, fh(3) = 1 2 , hf(3) = 2 3 , fhf(3) = 3 2 (check this!). In just one special case, the group maps a set of two points into itself, namely 1 2 (1 + i 3) and 1 2 (1 i 3). In just one special case, the group maps a set of three points into itself, namely 2, 1 2 and 1 (in fact there is another possibility if we include , namely 0, 1 and ). In these cases, the group can map the set of roots of certain quadratic or cubic equations into itself but in other cases there would have to be 6 roots for this to happen; the equation would have to be sextic at least.