For how many values \(a \in \mathbf{C}\), the equations \(x^2-8 x+7=0\) and \(x^2-2 a x+49=0\) have a common…
For how many values \(a \in \mathbf{C}\), the equations \(x^2-8 x+7=0\) and \(x^2-2 a x+49=0\) have a common root?
1
3
2
0
Solution
Let \(\alpha\) be the common root of given Equations
\(\begin{aligned}
\Rightarrow & \alpha^2-8 \alpha+7=0 \quad \ldots (i) \\
& \alpha^2-2 a \alpha+49=0 \quad \ldots (ii)
\end{aligned}\)
From Eq (i). - Eq. (ii)
\(\begin{aligned}
\Rightarrow \quad(-8+2 a) \alpha-42 & =0 \\
2(-4+a) \alpha & =42 \\
\alpha & =\frac{21}{a-4}
\end{aligned}\)
On putting value of \(\alpha\) in Eq. (i),
\(\begin{aligned}
\left(\frac{21}{a-4}\right)^2-8\left(\frac{21}{a-4}\right)+7 & =0 \\
441-168(a-4)+7(a-4)^2 & =0 \\
441-168 a+672+7\left(a^2+16-8 a\right) & =0 \\
7 a^2-224 a+1225 & =0
\end{aligned}\)
Here, \(D > 0\)
\(\therefore\) Above quadratic equation have two distinct real roots.
\(\therefore\) Number of possible value of \(a\) are 2.
Hence, option (c) is correct.