AP EAMCET · Maths · Determinants
If the system of simultaneous linear equations \(x+y-z=6,4 x+y+z=2\) and \(x+k y+z=-8\) has a unique solution \(x=2\), \(y=\beta, z=\gamma\), then the value of \(k\) satisfies the following quadratic equation
- A \(x^2-5 x+6=0\)
- B \(x^2+x-6=0\)
- C \(x^2-x-6=0\)
- D \(x^2+x-2=0\)
Answer & Solution
Correct Answer
(D) \(x^2+x-2=0\)
Step-by-step Solution
Detailed explanation
\(x+y-z=6\) \(\begin{aligned} & 4 x+y+z=2 \\ & x+k y+z=-8\end{aligned}\) \(x=2, y=\beta, z=\gamma\) \(y-3=4 \quad\) ...(i) \(y+z=-10\) ...(ii) \(2 y=-6\) \(\Rightarrow y=-3\) Put in eq no (i) \(3=-7\) Now, put values of \(x, y\) and \(z\), we get \(2+k(-3)+(-7)=-8\)…
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