Quadratic EquationHard
Question
Given that the equation $mx^{2} - 2(m + 2)x + m + 5 = 0,m \in R$ has no real root. Then the equation $(m - 6)x^{2} - 2(m + 2)x + (m + 5) = 0$ can have
Options
A.Two equal real roots
B.Two distinct real roots
C.No real roots
D.Exactly one real root
Solution
$mx^{2} - 2(m + 2)x + m + 5 = 0$ has no real roots
$$\Rightarrow \ m \neq 0\text{~and~}D < 0\ \Rightarrow \ m > 4$$
For the equation
$$\begin{matrix} (m - 6)x^{2} - 2(m + 2)x + m + 5 & \ = 0 \\ D & \ = 4(10m + 4) > 0\text{~for~}m > 4 \end{matrix}$$
⇒ Equation has two distinct real roots if $m \in (4,\infty) - \{ 6\}$ and one real root if $m = 6$
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