Question

(a) Prove that the roots of
${(a-b)}^2{x}^2+2(a+b-2c)x+1=0$ are real or imaginary according as c does not does lie between a and b,a<b.
(b) If the roots of the equation
$(m-3)x^2-2mx+5m=0$ are real and $+ive$, then prove that $mϵ]3,\frac{15}{4}]$
(c) If the equation $x^2+2(a+1)x+9a-5=0$ has only negative roots, then show that $a\ge 6$.
(d) If both the roots of the equation $x^2-6ax+2+2a+9a^2=0$ exceed $3$, then show that $a>\frac{11}{9}$.

Answer 1

(a) $B^2-4Ac=4{(a+b-2c)}^2-4{(a-b)}^2$
$\Delta =4\left[\right(2a-2c\left)\right(2b-2c\left)\right]$
by factors of $p^2-q^2$
$=4\times 2\times 2(c-a)(c-b)$, $a<b$.
If c lies between a and b, i.e. $a<c<b$, then
$\Delta =B^2-4AC$ is $-ive$
Hence the roots will be imaginary.
If c does not lie between a and b i.e. either $c<a$ or $c>b$ then
$\Delta =B^2-4AC$ is $+ive$
Hence the roots will be real.
(b) Roots are real and $+ive$ if
(1) $\Delta \ge 0$, (2) $S+ive$ (3) $P+ive$
(1) $\Rightarrow mϵ[0,\frac{15}{4}]$
(2) $\Rightarrow mϵ[0,\frac{15}{4}]m(m-3)>0$ or $m<0$ or $m>3$
(3) $\Rightarrow$ same as (2).
Hence combining,we can say
$m>3$ and $\le \frac{15}{4}$ $\le \frac{15}{4}$ or $mϵ]3,\frac{15}{4}]$.
(c) Since both the roots are $-ive$, the roots are real so that
$\Delta \ge 0,S<0,P>0$
$\Delta \ge 0\Rightarrow (a-1)(a-6)\ge 0$
or $a\le 1$ or $a\ge 6$.........(1)
$S<0\Rightarrow (a+1)<0$
$\Rightarrow$ $a+1>0\Rightarrow a>-1$.........(2)
$P>0\Rightarrow 9a-5>0\Rightarrow a>\frac{5}{9}$.........(3)
The value $a\ge 6$ satisfies all the three criteria.
(d) $f\left(x\right)=(x-\alpha )(x-\beta )=0$ must satisfy the following :
(a) $\Delta \ge 0$ (b) $S>6$
and $f\left(3\right)=(3-\alpha )(3-\beta )=(-)(-)=+ive$
$\Delta \ge 0\Rightarrow a\ge 1$.........(1)
$S>6\Rightarrow a>1$........(2)
$f\left(3\right)>0\Rightarrow 9a^2-20a+11>0$
or $9(a-1)(a-11/9)>0$
$\Rightarrow$ $a<1$ or $a>11/9$
Clearly $a>\frac{11}{9}$ satisfies both (1) and (2) also.