Question

Determine whether the given values are the solutions of the given equation or not.$x^2-(\sqrt{2}+\sqrt{3})x+\sqrt{6}=0;$ $x=\sqrt{2},x=\sqrt{3}$

• A. Only $x=\sqrt{2}$ is the solution of the equation
• B. Only $x=\sqrt{3}$ is the solution of the equation
• C. Both $x=\sqrt{2},\sqrt{3}$ are the solutions of the equation
• D. None of these

Answer 1

The correct option is C Both $x=\sqrt{2},\sqrt{3}$ are the solutions of the equation

$x^2-(\sqrt{2}+\sqrt{3})x+\sqrt{6}=0$
$x^2-\sqrt{2}x-\sqrt{3}x+\sqrt{6}=0$
$x(x-\sqrt{2})-\sqrt{3}(x-\sqrt{2})+0$
$(x-\sqrt{2})(x-\sqrt{3})=0$
Hence
$x=\sqrt{2}$ and $x=\sqrt{3}$ are solutions to the above equation.