Q155 / 270 · NDA 2011 (Phase 1) · Mathematics
If 5 and 7 are the roots of the equation
$\left|\begin{array}{ccc}
x & 4 & 5 \\
7 & x & 7 \\
5 & 8 & x
\end{array}\right|=0$
then what is the third root?
- Correct answerA-12
- B9
- C13
- D14
Solution
$\left|\begin{array}{lll}x & 4 & 5 \\ 7 & x & 7 \\ 5 & 8 & x\end{array}\right|=0$
$\begin{aligned} & \Rightarrow x\left(x^2-56\right)-7(4 x-40)+5(28-5 x)=0 \\ & \Rightarrow x^3-56 x-28 x+280+140-25 x=0 \\ & \Rightarrow x^3-109 x+420=0\end{aligned}$
Since 5 and 7 are the roots of the given equation so that
$x^2-12 x+35$ is a factor of the given equation, when we divide the given equation by $x^2-12 x+35$ it gives
$x+12=0$
$\Rightarrow \quad x=-12$.
Hence, the correct answer is option: 'A'.