Question

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

• A. $\frac{1}{2}(3+\sqrt{13}), \frac{1}{2}(3-\sqrt13)$
• B. $\frac{1}{2}(3-\sqrt{13}), \frac{-1}{2}(3+\sqrt13)$
• C. $\frac{1}{2}(3+\sqrt{17}), \frac{1}{2}(3-\sqrt17)$
• D. $\frac{1}{2}(3-\sqrt{17}), \frac{-1}{2}(3+\sqrt17)$

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
$\Rightarrow \, x= \, \frac{3 ± \sqrt{9+4}}{2}$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \bigg[ \because \, \, x= \frac{-b ± \sqrt{b^2-4ac}}{2a} \bigg]$
$\Rightarrow \, \, x= \frac{ 3 ± \sqrt{13}}{2}$
$\therefore $non-integer roots of given equation are
$\frac{1}{2}(3+\sqrt{13}), \frac{1}{2}(3-\sqrt{13})$