STEP II 1992 question 7 solution

Register

About Us | Help | Sign in

Create New Article | Editing Help

Advanced Search

Edit \| Hide
Welcome \| Study Help \| A Levels \| Applications, UCAS etc. \| University Discussion \| Careers \| General Discussion \| Have Your Say \| Health & Relationships \| Debate & Discussion \| Technology \| Sport \| Entertainment \| Chat \| Ask A Moderator

STEP II 1992 question 7 solution

From The Student Room

Jump to:navigation, search

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 7 solution

$x^3 - px^2 + qx - r = 0\\ \therefore abc = r, \quad ab+ac+bc = q,\quad a+b+c = p.$

$p=0\Rightarrow a+b+c=0\\ \text{Suppose } a=b, \text{ then the equation becomes} \\ a^2c = r,\quad a^2+2ac=q,\quad 2a+c=0.\\ 4q^3+27r^2 = 4(a^2+2ac)^3 + 27(a^2c)^2\\ =4(a^6 + 6a^5c + 12a^4c^2 + 8a^3c^3) + 27a^4c^2\\ = 4a^6 + 24a^5c + 48a^4c^2 + 32a^3c^3 + 27a^4c^2.\\ \text{But } c = 2a \Rightarrow 4q^3 + 27r^2 = 4a^6 - 24a^6 + 192a^6 - 256a^6 + 108a^6\\ =0 \text{ as required}.$

$\displaystyle 4\left( q - \frac{p^2}{3} \right)^2 + 27 \left( \frac{2p^3}{27} - \frac{pq}{3} + r\right)^2\\ = 4\left( a^2+2ac - \frac{4a^2+4ac+c^2}{3} \right)^2 + 27 \left[ \frac{16a^3 + 24a^2c + 12ac^2 + 2c^3}{27}$

$\displaystyle - \left \frac{2a^3 + 4a^2c + a^2c + 2ac^2}{3} + a^2c \right] ^2\\ = \frac{4(2ac-a^2-c^2)^3}{3^3} + 27\left( \frac{-2a^3 + 6a^2c - 6ac^2 + 2c^3}{27} \right) ^2\\ = \frac{4(-1)^3((a-c)^2)^3}{3^3} + \frac{27(-2)^2((a-c)^3)^2}{27^2}\\ = \frac{-4(a-c)^6}{27} + \frac{4(a-c)^6}{27} = 0.$