Question

(a) If the roots of the equation $x^2+a^2=8x+6a$ be real, then prove that a lies between $-2$ and $8$.
(b) Prove that if the roots of $9x^2+4ax+4=0$ are imaginary, then a must lie between $-3$ and $3$.
(c) The equation $x^2+2(m-1)x+m+5=0$ has at least one $+ive$ root. Determine the range for m.

Answer 1

(a) $B^2-4AC=\Delta =+ive$ for real
$\therefore$ $4(16+6a-a^2)=-4(a+2)(a-8)=+ive$
or $[a-(-2\left)\right](a-8)=-ive$ or $-2<a<8$.
Hence for roots to be real a should lie between $-2$ and $8$.
(b) $B^2-4AC=16a^2-36\times 4$
$=16(a^2-9)=16[a-(-3\left)\right](a-3)<0$
as the roots are imaginary. Hence a must lie between $-3$ and $3$.
(c) Atleast one $+ive$ root means :
(i) Both the roots are real (There cannot be one complex root as they occur in conjugate pairs)
(ii) Both are not $-ive$.
Condition (i) $\Rightarrow \Delta \ge 0$
or $4{(m-1)}^2-4(m+5)\ge 0$
or $m^2-2m+1-m-5\ge 0$
or $m^2-3m-4=+ive$
or $(m+1)(m-4)=+ive$
$\therefore$ $m\le -1,m\ge 4$ .......(1)
Both are not $-ive$ : Let us find the region when both are $-ive$.
$\therefore$ $S=-ive$ and $P=+ive$
$-2(m-1)=-ive$ and $m+5=+ive$
$m-1>0$ and $m+5>0$
$\therefore$ $m>1$ or $m>-5$
When $m>1$, it is automatically greater then $-5$.
Hence $m>1$ is the condition for both the roots to be $-ive$. Hence when both the roots are not $-ive$ i.e., at least one is $+ive$, we must have
$m\le 1$.........(2)
Plotting both (1) and (2) on real line and then taking their intersection i.e., common region we get the answer :
Condition (1) gives broken line region.
Condition (2) given dotted line region.
Common region is given by $m\le -1$
$\therefore$ $mϵ(-\infty ,-1)$.