Question

(a) Prove that the roots of
$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
are always real and they will be equal if and only if $a=b=c$.
(b) Examine the nature of the roots of the quadratic ${(b-x)}^2-4(a-x)(c-x)=0$ where a,b,c are real.
(c) Discuss the nature of the roots of the equation $x^2+2(3\lambda +5)x+2(9{\lambda }^2+25)=0$

Answer 1

(a) The given equation is
$3x^2-2x(a+b+c)+(ab+bc+ca)=0$
Roots Equal. $\Rightarrow B^2-4AC=0$
or $4{(a+b+c)}^2-12(ab+bc+ca)=0$
or $(a^2+b^2+c^2+2\sum ab-3\sum ab)=0$
or $(a^2+b^2+c^2-ab-bc-ca=0$
or $\frac{1}{2}[2a^2+2b^2+2c^2-2ab-2bc-2ca]=0$
or $\Delta =\frac{1}{2}[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]=0$
Clearly $\Delta \ge 0$ $\therefore$ Roots are real
They will be equal if $\Delta =0$. Hence
$a-b=0,b-c=0,c-a=0$ or $a=b=c$.
Converse : If $a=b=c$ then the given equation reduces to $3{(x-a)}^2=0$ which clearly has equal roots.
(b) The given equation can be written as
$3x^2-\left[4\right(a+c)-2b]x-(b^2-4ac)=0$
$\Delta ={\left[4\right(a+c)-2b]}^2+12(b^2-4ac)$
$={4\left[2\right(a+c)-b]}^2+({3b}^2-12ac)$
$=4\left[4\right(a^2+c^2+2ac)+b^2-4b(a+c)+3b^2-12ac]$
$=16[a^2+b^2+c^2-ab-bc-ca]$.........(1)
$=8[{(a-b)}^2+{(b-c)}^2+{(c-a)}^2]=+ive$
$\therefore$ Roots are real.
Note : we can also use the inequality
$a^2+b^2+c^2>ab+bc+ca$ in (1)
as $a^2+b^2>2ab,b^2+c^2>2bc,c^2+a^2>2ca$.
(c) $D=-4{(3\lambda -5)}^2$ If $\lambda \ne \frac{5}{3}$ their D is always $-ive$ and hence roots are complex. But if $\lambda =\frac{5}{3}$ their $D=0$ and is that case the root will be equal.