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If there are two values of ' $a$ ' which makes determinant
$$
\Delta=\left|\begin{array}{ccc}
1 & -2 & 5 \\
2 & a & -1 \\
0 & 4 & 2 a
\end{array}\right|=86
$$
Then, the sum of these number is
Options:
$$
\Delta=\left|\begin{array}{ccc}
1 & -2 & 5 \\
2 & a & -1 \\
0 & 4 & 2 a
\end{array}\right|=86
$$
Then, the sum of these number is
Solution:
2507 Upvotes
Verified Answer
The correct answer is:
-4
Given, $\Delta=\left|\begin{array}{ccc}1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a\end{array}\right|=86 \Rightarrow$
$$
\begin{aligned}
& \left|\begin{array}{ccc}
1 & -2 & 5 \\
2 & a & -1 \\
0 & 4 & 2 a
\end{array}\right|=86 \\
& \Rightarrow 1\left[2 a^2+4\right]+2[4 a+0]+5[8-0]=86 \\
& \Rightarrow \quad 2 a^2+4+8 a+40=86 \\
& \Rightarrow \quad \quad 2 a^2+8 a-42=0 \Rightarrow a^2+4 a-21=0 \\
& \Rightarrow \quad a^2+7 a-3 a-21=0 \Rightarrow(a+7)(a-3)=0 \\
& \therefore \text { Required sum }=-7+3=-4
\end{aligned}
$$
$$
\begin{aligned}
& \left|\begin{array}{ccc}
1 & -2 & 5 \\
2 & a & -1 \\
0 & 4 & 2 a
\end{array}\right|=86 \\
& \Rightarrow 1\left[2 a^2+4\right]+2[4 a+0]+5[8-0]=86 \\
& \Rightarrow \quad 2 a^2+4+8 a+40=86 \\
& \Rightarrow \quad \quad 2 a^2+8 a-42=0 \Rightarrow a^2+4 a-21=0 \\
& \Rightarrow \quad a^2+7 a-3 a-21=0 \Rightarrow(a+7)(a-3)=0 \\
& \therefore \text { Required sum }=-7+3=-4
\end{aligned}
$$
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