Question

The non integer roots of $x^4-3x^3-2x^2+3x+1=0$

• A. $\frac{1}{2}(3+\sqrt{13}),\frac{1}{2}(3-\sqrt{1}3)$
• B. $\frac{1}{2}(3-\sqrt{13}),\frac{-1}{2}(3+\sqrt{1}3)$
• C. $\frac{1}{2}(3+\sqrt{17}),\frac{1}{2}(3-\sqrt{1}7)$
• D. $\frac{1}{2}(3-\sqrt{17}),\frac{-1}{2}(3+\sqrt{1}7)$

Answer 1

Answer: (A)

Given $x^4-3x^3-2x^2+3x+1=0$
By using Hit & trial method, we have
(x - 1) is a factor of given equation
$\therefore$ (x - 1) (x³ - 2x² - 4x - 1) = 0
$\Rightarrow (x-1)[x^3-x^2-3x^2-3x-x-1]=0$
$\Rightarrow (x-1)[x^2(x+1)-3x(x+1)-1(x+1)]=0$
$\Rightarrow (x-1)(x+1)(x^2-3x-1)=0$
$\therefore$ x = 1, - 1 or x² - 3x - 1 = 0
Now x²-3x - 1 = 0
$\Rightarrow x=\frac{3\pm \sqrt{9+4}}{2}$
$[\because x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}]$
$\Rightarrow x=\frac{3\pm \sqrt{13}}{2}$
$\therefore$non-integer roots of given equation are
$\frac{1}{2}(3+\sqrt{13}),\frac{1}{2}(3-\sqrt{13})$