Question

If the root of the equation $x^2+a^2=8x+6a$ are real, then find the interval $a$ belongs to.

• A. $a\in [2,8]$
• B. $a\in [-2,8]$
• C. $a\in (2,8)$
• D. $a\in (-2,8)$

Answer 1

Answer: (B)

$a\in [-2,8]$

we know that for the equation $a{x}^2+bx+c=0$ to have real roots

$\Rightarrow b^2-4ac\ge 0$

so, for roots of equation $x^2+a^2=8x+6a$

to be real

$\Rightarrow (-8{)}^2-4\times 1(a^2-6a)\ge 0$

$\Rightarrow -4a^2+24a+64\ge 0$

$\Rightarrow a^2-6a-16\le 0$

$\Rightarrow (a-8)(a+2)\le 0$

(At $a<-2(a-8)(a+2)>0$)

$a\in [-2,8]$

($\therefore (a-8)$ and $(a+8)$ has odd power $\therefore$ sign change at $-2,8$)