Question

If $x^2-(a+b+c)x+(ab+bc+ca)=0$ has non real 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 triangle
• C. Nothing can be said
• D. None of these

Answer 1

The correct option is A Can be the sides of a triangle

$b^2-4ac<0$
Or
$(a+b+c{)}^2-4(ab+bc+ac)<0$
$a^2+b^2+c^2-2ab-2bc-2ac<0$
$(a-b-c{)}^2-4bc<0$
or
$(a-b-c{)}^2<4bc$
$(a-b-c)<2\sqrt{b}\sqrt{c}$
$a<b+c+2\sqrt{b}\sqrt{c}$
$a<(\sqrt{b}+\sqrt{c}{)}^2$
Or
${\sqrt{a}}^2<(\sqrt{b}+\sqrt{c}{)}^2$
$\sqrt{a}<\sqrt{b}+\sqrt{c}$
Hence sum of two sides is greater than the third side.
Hence a triangle can be formed with sides $\sqrt{a},\sqrt{b},\sqrt{c}$.