[5 marks sum]

Question 15 Marks

Solve the equation $x^4 + 2x^3 - 13x^2 + 2x + 1 = 0.$

Answer

Given equation
$x^4+2 x^3-13 x^2+2 x+1=0$
Dividing both sides by x2, we get
$x ^2+2 x -13+\frac{2}{x}+\frac{1}{x^2}=0$
$\Rightarrow\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-13=0$
Put $x+\frac{1}{x}=y$, squaring $x^2+\frac{1}{x^2}+2=y^2$
$\Rightarrow x^2+\frac{1}{x^2}=y^2-2$
Then $y ^2-2+2 y -13=0$
$\Rightarrow y^2+2 y-15=0$
$\Rightarrow y^2+5 y-3 y-15=0 $
$\Rightarrow y(y+5)-3(y+5)=0$
$\Rightarrow(y+5)(y-3)=0$
$\Rightarrow y+5=0 \text { or } y=-5$
or $y-3=0$ or $y=3$
But $x+\frac{1}{x}=-5$
Then $x+\frac{1}{x}=-5$
$\Rightarrow x^2+1=-5 $
$\Rightarrow x^2+5 x+1=0 $
$\Rightarrow x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$x=\frac{-5 \pm \sqrt{25-4}}{2 \times 1} $
$x=\frac{-5 \pm \sqrt{25-4}}{2} $
$x=\frac{-5 \pm \sqrt{21}}{2}$
$\text { or } x+\frac{1}{x}=3 $
$\Rightarrow x^2+1=3 $
$\Rightarrow x^2-3 x +1=0 $
$\Rightarrow x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$x =\frac{-(-3) \pm \sqrt{9-4}}{2} $
$x =\frac{3 \pm \sqrt{5}}{2}$
Hence $x=\frac{-5 \pm \sqrt{21}}{2}, \frac{3 \pm \sqrt{5}}{2}$

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Question 25 Marks

Solve for x:
$\left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0$

Answer

Given equation
$\left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0$
Put $x-\frac{1}{x}=y$, squaring $x^2+\frac{1}{x^2}-2=y^2$
$\Rightarrow x^2+\frac{1}{x^2}=y^2+2$
Then given rquation becomes :
$x^2+\frac{1}{x^2}+2-\frac{3}{2}\left(x-\frac{1}{x}\right)-4=0 $
$\Rightarrow y^2+2+2-\frac{3}{2} y-4=0 $
$\Rightarrow y^2+4-\frac{3}{2} y-4=0 $
$ \Rightarrow 2 y^2-3 y=0 $
$\Rightarrow y(2 y-3)=0 $
$\Rightarrow y=0 \text { or } 2 y-3=0 \text { i.e., } y=\frac{3}{2}$
$\text { But } x-\frac{1}{x}=y$
$\text { Then } x-\frac{1}{x}=0$
$ \Rightarrow x 2-1=0$
$ \Rightarrow x = \pm 1 $
$ \text { or } $
$x-\frac{1}{x}=\frac{3}{2} $
$\Rightarrow \frac{x^2-1}{x}=\frac{3}{2}$
$\Rightarrow 2 x^2-2=3 x$
$\Rightarrow 2 x^2-3 x-2=0$
$ \Rightarrow 2 x^2-4 x+x-2=0 $
$ \Rightarrow 2 x(x-2)+1(x-2)=0$
$\Rightarrow(x-2)(2 x+1)=0 $
$ \Rightarrow x-2=0 \text { or } 2 x+1=0$
$\Rightarrow x=2 \text { or } x=-\frac{1}{2} $
$\text { Hence, } x= \pm 1,2,-\frac{1}{2} .$

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Mathematics [5 marks sum]

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