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If 5 and 7 are the roots of the equation
$\left|\begin{array}{lll}x & 4 & 5 \\ 7 & x & 7 \\ 5 & 8 & x\end{array}\right|=0$, then what is the third root?
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$\left|\begin{array}{lll}x & 4 & 5 \\ 7 & x & 7 \\ 5 & 8 & x\end{array}\right|=0$, then what is the third root?
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Verified Answer
The correct answer is:
$-12$
Given, $\left|\begin{array}{lll}x & 4 & 5 \\ 7 & x & 7 \\ 5 & 8 & x\end{array}\right|=0$
$\Rightarrow x\left(x^{2}-56\right)-4(7 x-35)+5(56-5 x)=0$
$\Rightarrow x^{3}-56 x-28 x+140+280-25 x=0$
$\Rightarrow x^{3}-109 x+420=0$
$\Rightarrow(x-5)(x-7)(x+12)=0$
$\Rightarrow x=-12$
Hence, the third root is $-12$.
$\Rightarrow x\left(x^{2}-56\right)-4(7 x-35)+5(56-5 x)=0$
$\Rightarrow x^{3}-56 x-28 x+140+280-25 x=0$
$\Rightarrow x^{3}-109 x+420=0$
$\Rightarrow(x-5)(x-7)(x+12)=0$
$\Rightarrow x=-12$
Hence, the third root is $-12$.
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