Question

Number of real roots of $x+\sqrt{1-x^2}=5$

Answer 1

Solution:-

$x+\sqrt{1-x^2}=5\left\{Given\right\}$

$\Rightarrow x-5=-\sqrt{1-x^2}$

Squaring both sides, we have

${(x-5)}^2={(-\sqrt{1-x^2})}^2$

$x^2+25-2\times 5\times x=1-x^2$

$\Rightarrow x^2+25-10x=1-x^2$

$\Rightarrow 2x^2-10x+24=0$

$\therefore D=b^2-4ac={(-10)}^2-4\times 2\times 24=-92$

Since $D<0$, there will be no real roots.