ABC26GN1491 · Square Roots and Cube Roots

$a=\frac{\sqrt{3}}{2}$ (given) $$\begin{aligned} & \therefore \sqrt{1+a}+\sqrt{1-a} \\ & =\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}} \\ & =\sqrt{\frac{2+\sqrt{3}}{2}}+\sqrt{\frac{2-\sqrt{3}}{2}} \end{aligned}$$ $=\sqrt{\frac{2(2+\sqrt{3})}{4}}+\sqrt{\frac{2(2-\sqrt{3})}{4}}$ $=\sqrt{\frac{4+2 \sqrt{3}}{4}}+\sqrt{\frac{4-2 \sqrt{3}}{4}}$ $=\sqrt{\frac{3+1+2 \times \sqrt{3} \times 1}{2}}+\sqrt{\frac{3+1-2 \times \sqrt{3} \times 1}{2}}$ $$\left\{\begin{array}{l} \sqrt{3}^{2}+(1)^{2}-2 \times \sqrt{3} \times 1=(\sqrt{3}-1)^{2} \\ \because \quad(\sqrt{3})^{2}+(1)^{2}+2 \times \sqrt{3} \times 1=(\sqrt{3}+1)^{2} \\ a^{2}+b^{2}+2 a b=(a+b)^{2} \\ a^{2}+b^{2}-2 a b=(a-b)^{2} \end{array}\right\}$$ $=\frac{\sqrt{(\sqrt{3}+1)^{2}}}{2}+\frac{\sqrt{(\sqrt{3}-1)^{2}}}{2}$ $=\frac{\sqrt{3}+1+\sqrt{3}-1}{2}$ $=\frac{2 \sqrt{3}}{2}=\sqrt{3}$