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Express $x - \frac{1}{x} = 3$ as a quadratic equation in standard form and hence find its roots. Also, find the value of 'a' for which the equation $x + \frac{1}{x} = a$, when expressed as a quadratic equation, has real and equal roots.
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$x - \frac{1}{x} = 3 \implies x^2 - 1 = 3x \implies x^2 - 3x - 1 = 0$ (1 Mark)
n$D = (-3)^2 - 4(1)(-1) = 9 + 4 = 13$ (1 Mark)
nRoots are $\frac{3 + \sqrt{13}}{2}$, $\frac{3 - \sqrt{13}}{2}$ (1 Mark)
nNow, $x + \frac{1}{x} = a \implies x^2 - ax + 1 = 0$ (1/2 Mark)
nSince roots are real and equal, $D = 0$
n$(-a)^2 - 4(1)(1) = 0 \implies a^2 - 4 = 0$ (1/2 Mark)
n$a^2 = 4 \implies a = 2$ or $a = -2$ (1/2+1/2 Mark)
n$D = (-3)^2 - 4(1)(-1) = 9 + 4 = 13$ (1 Mark)
nRoots are $\frac{3 + \sqrt{13}}{2}$, $\frac{3 - \sqrt{13}}{2}$ (1 Mark)
nNow, $x + \frac{1}{x} = a \implies x^2 - ax + 1 = 0$ (1/2 Mark)
nSince roots are real and equal, $D = 0$
n$(-a)^2 - 4(1)(1) = 0 \implies a^2 - 4 = 0$ (1/2 Mark)
n$a^2 = 4 \implies a = 2$ or $a = -2$ (1/2+1/2 Mark)