Question

If $x^2-(a+b+c)x+(ab+bc+ca)=0$ has imaginary roots, where $a,b,c\in R^{+},$ then $\sqrt{a},\sqrt{b},\sqrt{c}$

• A. can be the sides of a triangle
• B. cannot be the sides of a triangle
• C. nothing can be said
• D. none of these

Answer 1

Answer: (A)

can be the sides of a triangle

Since the given equation has imaginary roots

$\Rightarrow D<0\Rightarrow {(a+b+c)}^2-4(ab+bc+ca)<0$

$\Rightarrow (a^2+b^2+c^2-2ab-2bc+2ac)<4ac$

$\Rightarrow {(-a+b-c)}^2<4ac$ $\Rightarrow -2\sqrt{ac}<a-b+c$

$\Rightarrow (a+c+2\sqrt{ac})>b$ $\Rightarrow {(\sqrt{a}+\sqrt{c})}^2>b\Rightarrow \sqrt{a}+\sqrt{c}>\sqrt{b}.$

Similarly, $\sqrt{b}+\sqrt{c}>\sqrt{a}$ and $\sqrt{a}+\sqrt{b}>\sqrt{c}.$

Therefore,$\sqrt{a},\sqrt{b},\sqrt{c}$ can be the sides of a triangle.