(a) If \(\frac{\sqrt{5} + 4}{3 - 2\sqrt{5}} - \frac{2 + \sqrt{5}}{4 - 2\sqrt{5}} = a + b\sqrt{5}\), find the values of a and b. (b)(i) Evaluate : \(\begin{v...

Answer Details

(a) Simplify \(\dfrac{\sqrt5+4}{3-2\sqrt5}-\dfrac{2+\sqrt5}{4-2\sqrt5}=a+b\sqrt5\).

Rationalise the first fraction by \(3+2\sqrt5\) (denominator \(9-20=-11\)):

\[\frac{(\sqrt5+4)(3+2\sqrt5)}{-11}=\frac{22+11\sqrt5}{-11}=-2-\sqrt5\]

Rationalise the second by \(4+2\sqrt5\) (denominator \(16-20=-4\)):

\[\frac{(2+\sqrt5)(4+2\sqrt5)}{-4}=\frac{18+8\sqrt5}{-4}=-\frac{9}{2}-2\sqrt5\]

Subtract:

\[(-2-\sqrt5)-\left(-\frac{9}{2}-2\sqrt5\right)=-2+\frac{9}{2}+(-\sqrt5+2\sqrt5)=\frac{5}{2}+\sqrt5\]

Hence \(a=\dfrac{5}{2},\ b=1\).

(b)(i) Evaluate the determinant.

\[\begin{vmatrix}2&-1&2\\1&3&4\\1&2&1\end{vmatrix}=2(3-8)-(-1)(1-4)+2(2-3)=-10-3-2=-15\]

(ii) Writing the system as \(2x-y+2z=-5,\ x+3y+4z=1,\ x+2y+z=-2\), the coefficient determinant is \(-15\). By Cramer's rule, replace the x-column with the constants:

\[D_x=\begin{vmatrix}-5&-1&2\\1&3&4\\-2&2&1\end{vmatrix}=-5(3-8)+1(1+8)+2(2+6)=25+9+16=50\]\[x=\frac{D_x}{D}=\frac{50}{-15}=-\frac{10}{3}\approx -3.33\]