Question

(a) a,b,c are lengths of sides of an scalene triangle. If equation
$x^2+2(a+b+c)x+3\lambda (ab+bc+ac)=0$
has real and distinct roots, then the value of $\lambda$ is given by :
(a) $\lambda <\frac{4}{3}$
(b) $\frac{11}{3}<\lambda <\frac{17}{3}$
(c) $\frac{11}{6}<\lambda <\frac{15}{4}$
(d) $\lambda \ge 1$
(b) If a,b are the roots of $x^2-10cx-11d=0$ and c,d are the roots of $x^2-10ax-11b=0$, then find the value, of $a+b+c+d$. (a,b,c,d are distinct numbers).

Answer 1

A quadratic equation has real and distinct roots implies its $\Delta \ge 0$.
or $4{(a+b+c)}^2-12\lambda (ab+bc+ca)\ge 0$
or $(\sum a^2+2\sum ab)-3\lambda \sum ab\ge 0$
or $\sum a^2\ge (3\lambda -2)\sum ab$ or $3\lambda -2<\frac{\sum a^2}{\sum ab}$.......(1)
Now in any triangle sum of two sides is greater than the third.
$\therefore$ $a+b>c$ $b+c>a$, $c+a>b$
or $a>c-b$, $b>a-c$, $c>b-a$
or $a^2+b^2+c^2>2(a^2+b^2+c^2)-2\sum ab$
or $2\sum ab>\sum a^2$ or $\frac{\sum a^2}{\sum ab}<2$......(2)
Hence from (1) and (2).
$3\lambda -2<2$ or $\lambda <4/3$.
(b) Ans.1210
Given $a+b=10c$, $ab=11d$.........(1)
$c+d=10a$, $cd=-11b$........(2)
$\therefore$ $a+b+c+d=10(a+c)$......(3)
So we have to find the value of $a+c$ so that by (3)
We get the value of $a+b+c+d$.
From relations (1) and (2) on adding,we have
$b+d=9(a+c)$........(4)
and multiplying relations in (1) and (2)
$\left(ac\right)bd=121bd$ $\Rightarrow$ $ac=121$.........(5)
Multiplying (1) by a and (2) by c and adding,
$a^2+c^2+(ab+cd)=(10+10)ac$
or $a^2+c^2-11(b+d)=20ac$ by (1),(2),
or ${(a+c)}^2-11\times 9(a+c)=22ac$
or $t^2-99t-22\left(121\right)=0$ by (4) and (5)
or $t^2-(121-22)t-22\left(121\right)=0$
$(t-121)(t+22)=0$ $\therefore$ $t=121=a+c$
Hence from (3),
$a+b+c+d=10(a+c)=10\times 121=1210$