First slide

Theory of equations

Question

If the roots of the equation $x^{2} - 8 x + (a^{2} - 6 a) = 0$ are real, then

Moderate

A

$- 2 < a < 8$
B

$2 < a < 8$
C

$- 2 \leq a \leq 8$
D

$2 \leq a \leq 8$

Solution

Roots of $x^{2} - 8 x + (a^{2} - 6 a) = 0$ are real. So $D \geq 0$
$\Rightarrow 64 - 4 (a^{2} - 6 a) \geq 0 \Rightarrow 16 - a^{2} + 6 a \geq 0 \Rightarrow a^{2} - 6 a - 16 \leq 0 \Rightarrow (a - 8) (a + 2) \leq 0$
Now we have two cases:
Case I: $(a - 8) \leq 0 and (a + 2) \geq 0$
$\Rightarrow a \leq 8 and a \leq - 2$
Case II: $(a - 8) \geq 0 and (a + 2) \leq 0$
$\Rightarrow a \geq 8 and a \leq - 2$ but it is impossible
Therefore, we get $- 2 \leq a \leq 8$

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