Question

If the equation $a{x}^2+bx+c=0,$ where $a,b,c$ are the sides of a $△ABC,$ and the equation $x^2+\sqrt{2}x+1=0$ have a common root, then $\angle C=$

• A. ${90}^{\circ }$
• B. ${35}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${45}^{\circ }$

Clearly, the roots of $x^2+\sqrt{2}x+1=0$ are non-real complex.
So, the one root common implies both roots are common.
So, $\frac{a}{1}=\frac{b}{\sqrt{2}}=\frac{c}{1}=k$
$\therefore \cos C=\frac{a^2+b^2-c^2}{2ab}$
$=\frac{k^2+2k^2-k^2}{2\sqrt{2}{k}^2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \angle C={45}^{\circ }$