Problems

Filters
Clear Filters

3 problems found

2019 Paper 1 Q4
D: 1500.0 B: 1500.0
polynomials nested radicals quartic factorisation quadratic formula surds completing the square simultaneous equations

  1. Find integers \(m\) and \(n\) such that $$\sqrt{3+2\sqrt{2}} = m + n\sqrt{2}.$$
  2. Let \(f(x) = x^4 - 10x^2 + 12x - 2\). Given that the equation \(f(x) = 0\) has four real roots, explain why \(f(x)\) can be written in the form $$f(x)=(x^2 + sx + p)(x^2 - sx + q)$$ for some real constants \(s\), \(p\) and \(q\), and find three equations for \(s\), \(p\) and \(q\). Show that $$s^2(s^2 - 10)^2 + 8s^2 - 144 = 0$$ and find the three possible values of \(s^2\). Use the smallest of these values of \(s^2\) to solve completely the equation \(f(x) = 0\), simplifying your answers as far as you can.

Solution:

  1. \((1+\sqrt{2})^2 = 3 + 2\sqrt{2}\) so \(\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}\)
  2. We can always factorise any quartic in the form \((x^2+ax+b)(x^2+cx+d)\), since \(x^3\) has a coefficient of \(a+b\) we must have \(a = -b\), ie the form in the question. \begin{align*} && 0 &= (x^2+sx+p)(x^2-sx+q) \\ &&&= x^4+(p+q-s^2)x^2+s(q-p)x+pq \\ \Rightarrow && pq &= -2 \\ && s(q-p) &= 12 \\ && p+q-s^2 &= -10 \\ \\ && p+q &= s^2-10 \\ && (p+q)^2 &= (s^2-10)^2 \\ && (q-p)^2 &= \frac{12}{s^2} \\ \Rightarrow && (s^2-10)^2 &= \frac{12}{s^2} + 4pq \\ && (s^2-10)^2 &= \frac{144}{s^2} -8 \\ && 0 &= s^2(s^2-10)^2+8s^2-144 \\ &&&= s^6-20s^4+108s^2-144 \\ &&&= (s^2-2)(s^2-6)(s^2-12) \end{align*} Suppose \(s = \sqrt{2}\), and we have \begin{align*} && q-p &= 6\sqrt{2} \\ && p+q &= -8 \\ \Rightarrow && q &= 3\sqrt{2}-4 \\ && p &= -4-3\sqrt{2} \end{align*} Solving our quadratic equations, we have \begin{align*} && 0 &= x^2-\sqrt{2}x-4+3\sqrt{2} \\ \Rightarrow && x &= \frac{\sqrt{2}\pm \sqrt{2-4\cdot(-4+3\sqrt{2})}}{2} \\ &&&= \frac{\sqrt{2}\pm \sqrt{18-12\sqrt{2}}}{2} \\ &&&= \frac{\sqrt{2}\pm (2\sqrt{3}-\sqrt{6})}{2} \\ \\ && 0 &= x^2+\sqrt{2}x-3\sqrt{2}-4 \\ && x &= \frac{-\sqrt{2} \pm \sqrt{2-4\cdot(3\sqrt{3}-4)}}{2}\\ && &= \frac{-\sqrt{2} \pm \sqrt{18+12\sqrt{2}}}{2}\\ && &= \frac{-\sqrt{2} \pm (\sqrt{6}+2\sqrt{3})}{2}\\ \end{align*}

View
2007 Paper 1 Q6
D: 1500.0 B: 1489.2
polynomials difference of powers factorisation algebraic manipulation integer solutions substitution quadratic formula number theory

  1. Given that \(x^2 - y^2 = \left( x - y \right)^3\) and that \(x-y = d\) (where \(d \neq 0\)), express each of \(x\) and \(y\) in terms of \(d\). Hence find a pair of integers \(m\) and \(n\) satisfying \(m-n = \left( \sqrt {m} - \sqrt{n} \right)^3\) where \(m > n > 100\).
  2. Given that \(x^3 - y^3 = \left( x - y \right)^4\) and that \(x-y = d\) (where \(d \neq 0\)), show that \(3xy = d^3 - d^2\). Hence show that \[ 2x = d \pm d \sqrt {\frac{4d-1 }{3}} \] and determine a pair of distinct positive integers \(m\) and \(n\) such that \(m^3 - n^3 = \left( m - n \right)^4\).

Solution:

  1. \(\,\) \begin{align*} && x^2-y^2 &=(x-y)^3 \\ \Rightarrow && x+y &=d^2 \\ && x-y &= d \\ \Rightarrow && x &= \tfrac12(d^2+d) \\ && y &= \tfrac12(d^2-d) \end{align*} Therefore consider \(x^2 = m, y^2 = n\), so \(m = \tfrac14(d^2+d)^2, n = \tfrac14(d^2-d)^2\) so we want \(d^2-d > 20\), so \(d = 6, n = 225, m = 441\).
  2. \(\,\) \begin{align*} && x^3-y^3 &= (x-y)^4 \\ \Rightarrow && x^2+xy+y^2 &= (x-y)^3 \\ && d^3 &= (x-y)^2+3xy \\ && d^3 &= d^2 + 3xy \\ \Rightarrow && 3xy &= d^3 - d^2 \\ \Rightarrow && 3x(x-d) &= d^3-d^2 \\ \Rightarrow && 0 &= 3x^2-3dx-(d^3-d^2) \\ \Rightarrow && 2x &=d \pm \sqrt{d^2+4\frac{(d^3-d^2)}{3}} \\ &&&= d \pm d \sqrt{\frac{3+4d-4}{3}} \\ &&&= d \pm d \sqrt{\frac{4d-1}{3}} \end{align*} Therefore we need \(\frac{4d-1}{3}\) to be an odd square. \(y = x-d = -\frac{d}{2} \pm \frac{d}{2} \sqrt{\frac{4d-1}{3}}\). Since we want positive values, we should take the positive square roots. \(d = \frac{3 \cdot 3^2 + 1}{4} = 7\) we have \(2x = 7 +7 \cdot 3 = 28 \Rightarrow x = 14, y = 7\)

View
1996 Paper 1 Q5
D: 1484.0 B: 1500.0
complex numbers surds quadratic equation equating real and imaginary parts rationalization quadratic formula

  1. Find all rational numbers \(r\) and \(s\) which satisfy \[ (r+s\sqrt{3})^{2}=4-2\sqrt{3}. \]
  2. Find all real numbers \(p\) and \(q\) which satisfy \[ (p+q\mathrm{i})^{2}=(3-2\sqrt{3})+2(1-\sqrt{3})\mathrm{i}. \]
  3. Solve the equation \[ (1+\mathrm{i})z^{2}-2z+2\sqrt{3}-2=0, \] writing your solutions in as simple a form as possible.
{[}No credit will be given to answers involving use of calculators.{]}

Solution:

  1. Suppose \begin{align*} && 4 - 2\sqrt{3} &= (r+s\sqrt{3})^2 \\ &&&= r^2+3s^2+2sr \sqrt{3} \\ \Rightarrow && rs &= -1 \\ && r^2+3s^2 &= 4 \\ \Rightarrow && (r,s) &= (1,-1), (-1,1) \end{align*}
  2. \begin{align*} && (3-2\sqrt{3})+2(1-\sqrt{3})i &= (p+qi)^2 \\ &&&= p^2-q^2 + 2pq i \\ \Rightarrow && pq &= (1-\sqrt{3}) \\ && p^2 - q^2 &= 3-2\sqrt{3} \\ \Rightarrow &&3-2\sqrt{3} &= p^2 - \frac{(1-\sqrt{3})^2}{p^2} \\ \Rightarrow && 0 &= p^4-(3-2\sqrt{3})p^2-(4-2\sqrt{3}) \\ &&&= (p^2-(4-2\sqrt{3}))(p^2+1) \\ \Rightarrow && p &= \pm (1-\sqrt{3}) \\ && q &=\mp \frac12(1+\sqrt{3}) \end{align*}
  3. \begin{align*} && 0 &= (1+i)z^2 - 2z + 2(\sqrt{3}-1) \\ \Rightarrow && z &= \frac{2 \pm \sqrt{4-4(1+i)2(\sqrt{3}-1)}}{2(1+i)} \\ &&&= \frac{1 \pm \sqrt{1-(1+i)2(\sqrt{3}-1)}}{1+i} \\ &&&= \frac{1 \pm \sqrt{(3-2\sqrt{3})+(2-2\sqrt{3})i}}{1+i} \\ &&&= \frac{1 \pm (1 - \sqrt{3}) \mp \frac12 (1+\sqrt{3})i}{1+i} \\ &&&= \frac{5-\sqrt{3}}{4} + \frac{3-3\sqrt{3}}{4}i, \\ &&& \frac{\sqrt{3}-1}{4} + \frac{1+3\sqrt{3}}{4}i \end{align*}

View