Preliminary Maths Extension
Question If, and are the roots of x 5x x 0, find the following. (d) (e) Question If p, q and r are the roots of x x x 4 0, evaluate the following. pq r pq qr rp p q q r r p Question If,, and are the roots of the equation.., evaluate the following.
Question 4 If, and are the roots of x 6x x 0, find the value of the following. (d) Question 5 If, and are the roots of x 6x x 0, find the following. (d) (e) Question 6 If,, and are the roots of 4 x x x x 4 0, find the following. (d) (e)
Question 7 If,, and are the roots of 4 x x x 5 0, find the following. Question 8 If x x x k 5 7 0 has a double integral root, then find k. Question 9 If two roots of x bx cx 0 are and, find the third root. Question 0 Form a cubic equation whose roots are,,. Question The product of two roots of the equation x kx 8x 4 0 is. Find the value of k. Question The roots of x ax bx c 0 are 4, and. What are the values of a, b and c? Question If x x ax 8 0 has one root equal to the sum of the other two, find the value of a. Question 4 Solve x x x 9 8 0 if the roots are in arithmetic progression.
4 Question 5 Find the roots of x x x 80 0, if they are in arithmetic progression. Question 6 Solve x x x 4 56 64 0 if the roots are in geometric progression. Question 7 For what values of k does the quadratic equation k x kx k differing by? 4 4 0 have two roots Question 8 The roots of x x x m 6 0 are in arithmetic progression. Find m. Solve the equation. Question 9 The equation 0 has two equal roots. Prove that. L 4H. x Hx L Question 0 The equation x kx x 4 8 0 has two roots of equal size but opposite sign. Find the value of k Solve the equation.
5 Question The polynomial P x x x kx 4 has roots,,. Find the value of. Find the value of. It is known that two of the roots are equal in magnitude but opposite in sign. Find the third root and hence find the value of k. Question Let Px x x Qx ax b, where numbers. When When Px is divided by x the remainder is -. Px is divided by x the remainder is. What is the value of b? What is the remainder when P x is divided by x x Q x is a polynomial and a and b are real?
6 Fully Worked Solutions Question d = a c a b 5 a (d) (e) 5 Question p q r b 4 a pq qr rp c a p q q r r p p q r pr qr rp p q r pq qr rp 6 0 Question b a c 4 a 4 Question 4 b 6 a c a (d) Question 5
7 b 6 a c a (d) 6 (e) Question 6 4 x 4x x x 0 b 4 4 a c a d a (d) e a (e) Question 7 4 x x 5x 0 e a d 5 5 a 5 5 Question 8 let the roots be, and 5 7 k substitute 5 7 0 4 7 0 0 7 0 0 7 0 7 0 5 into 7 5
8 7 or is the only solution, as is integral, substitute into x 5x 7x k 0 5 7 k 0 k Question 9 the roots are, and d a Question 0 x x x 0 x x x 0 x x x 0 Question the product of two roots is, d 4 a substitute into substitute x into x kx 8x 4 0 k 8 4 0 k Question x 4 x x 0 x 5x x 8 0 a 5, b, c 8 Question 8 8 4 substitute x 4 into x 8x ax 0 4 84 a 4 0 a 9 Question 4 x x 9x 8 0 let the three roots be d,, d then d d 4 d d 8
9 d d 4 d4 d 7 4 4 4 8 6 d 7 d 9 d x 4,4,4 x,4,7 Question 5 Since the roots are in AP, let the roots be: a d, a, a d The sum of the roots is a d a a d, giving a 4. Thus one root is x 4 On dividing the LHS of the equation by x 4, we have: x 4x 8x 0 0 x 8x 0 0 gives x0 x 0, so the three roots are x, 4, 0 Question 6 let the three roots be,, r, then r 64 4 r r 56.. r 56 r 7 r r r 5r 0 r r 0 r or x,4,8 Question 7 Let and be the roots of k 4 x kx 4k 0 k 4k k 4 4 4 k 4 9k 6k Substituting from,, and 4 : 9k k 4 6 k k 4 k k6 0 k k 0 k or Question 8 k 4 k 4
0 x 6x x m 0 Since the roots are in AP, let them be a d, a, a d Then a d a a d 6 a 6, a, So one root is x The root x satisfies the given equation, 8 4 6 m 0 m 0 The equation is x 6x x 0 0 d d 0 d d d d 0 5 4 d 5 d 9 d therefore a d, a, a d,,5 Question 9 x Hx L 0 Let, and be the roots 0 and L Substituting in L : L L () satisfies (), so H L 0 L Using (): H L 0 L L L L This gives or. Using (): Hence L 4H. Obviously H 0. H 8H 8H Question 0 Since two of the roots are equal in size and opposite in sign, let the roots be,,. k kx 4x 8 0...() The sum of the roots: k, and k The root satisfies the equation, k k 4k 8 0 k Substituting k in (): x x 4x 8 0 x x 4 x 0 x x 4 0 The required roots are -, -, Alternatively: k 8 4, then Question If Px x x kx 4 has roots,, then 4 4
If two of the roots are equal in magnitude, but opposite in sign, let the root, and hence the roots are,, From, Thus the third root is. Also from, a 4 a 4 a a That is, the roots of P P P 0 P x are,, P k 4 88k 4 k 4 0 k Question P Q a b 0 0 b b But P and thus b when P x is divided by Now x, the remainder is, i.e. P P Q a b 0 4a b 4a b Substituting b in 4ab gives 4a a a x b, thus substitute a and b, the remainder is then we get x x 8