Question

If the roots of the equation $x^2-8x+a^2-6a=0$ are real, then the value of $a$ will be

• A. $-2<a<8$
• B. $-2\le a\le 8$
• C. $2<a<8$
• D. $2\le a\le 8$

Answer 1

Answer: (B)

$-2\le a\le 8$

Roots of the equation $x^2-8x+a^2-6a=0$ are real if the discriminant, $B^2-4AC\ge 0$
$where,A=1,B=-8,C=a^2-6a$
$B^2-4AC\ge 0$
$\Rightarrow 64-4(a^2-6a)\ge 0$
$\Rightarrow -4(a^2-6a)+64\ge 0$
$\Rightarrow -4(a^2-6a-16)\ge 0$
Divide both sides by $-4$,we get (The sign of inequality will change if we multiply or divide by any real negative number)
$\Rightarrow a^2-6a-16\le 0$
$\Rightarrow a^2-8a+2a-16\le 0$
$\Rightarrow a(a-8)+2(a-8)\le 0$
take common out $(a-8)$ from both the terms
$\Rightarrow (a+2)(a-8)\le 0$
$\therefore -2\le a\le 8$
Hence, option B.