# Find the range of $k$.

Let a, b, c be the sides of a triangle where $a\neq c$ and $k \in R$. If the roots of the equation $x^2+ 2(a + b +c)x + 3k(ab + bc + ca) = 0$ are real, then find the interval in which $k$ lies.

I have used the fact that equation has real roots, but how to use the fact a,b,c, are sides of a triangle.

• where is $k$ in your equation? – Sandeep Thilakan Mar 23 '14 at 7:35

Since $a,b,c$ are sides of a triangle,

$$|a-b|<c, \quad |b-c|<a, \quad |c-a|<b$$

Squaring and adding these inequalities, we get

$$a^2+b^2+c^2-2ab-2bc-2ca < 0$$

i.e. $$\dfrac{a^2+b^2+c^2}{ab+bc+ca} < 2$$

From the discriminant equation for the quadratic, we have $$k \leq \dfrac{a^2+b^2+c^2}{3(ab+bc+ca)} + \dfrac{2}{3}$$

and hence $$k < \dfrac{4}{3}$$

$$(2(a+b+c))^2\ge4(1)(3k(ab+bc+ca)$$ (since if $ax^2+bx+c=0$ then $b^2\ge4(a)(c)$ for real roots to exist) $$k\le\frac{4(a+b+c)^2}{12(ab+bc+ca)}$$

• I believe it should actually be $$k \leq \frac{(a+b+c)^2}{3 (ab + bc + ac)} = \frac{1}{3} \left[\frac{a^2 + b^2 + c^2}{(ab+bc+ca)} + 2 \right] .$$ – izœc Mar 23 '14 at 9:12