Question

Find the number of real roots of the equation $\left|\begin{array}{ccc}0& x-a& x-b\\ x+a& 0& x-c\\ x+b& x+c& 0\end{array}\right|=0$, $a\ne b\ne c,b(a+c)>ac$

• A. 1
• B. 0
• C. 3
• D. None of these.

Answer 1

The correct option is C 3

$\left|\begin{array}{ccc}0& x-a& x-b\\ x+a& 0& x-c\\ x+b& x+c& 0\end{array}\right|=0$
expanding along first column,

$(x+a)(x-b)(x+c)+(x+b)(x-a)(x-c)=0$

$[x^3+(a-b+c)x^2+(ac-ab-bc)x-abc]+[x^3+(b-a-c)x^2+(ac-ab-bc)x+abc]=0$

$x^3+(ac-ab-bc)x=0\Rightarrow x=0,\pm \sqrt{b(a+c)-ac}$,

Since given $\left[b\right(a+c)-ac>0]$

Hence, there exist 3 real roots.