Answer with explanation:
Ques 1)
\(x=\dfrac{1}{\sqrt{3}-\sqrt{2}}\)
Now we are asked to find the value of:
\(\sqrt{x}-\dfrac{1}{\sqrt{x}}\)
We know that:
\((\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=x+\dfrac{1}{x}-2\)
Also:
\(x=\dfrac{1}{\sqrt{3}-\sqrt{2}}\) could be written as:
\(x=\dfrac{1}{\sqrt{3}-\sqrt{2}}\times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\\\\\x=\dfrac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}\)
since, we know that:
\((a+b)(a-b)=a^2-b^2\)
Hence,
\(x=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}\\\\\\x=\sqrt{3}+\sqrt{2}\)
Also,
\(\dfrac{1}{x}=\sqrt{3}-\sqrt{2}\)
Hence, we get:
\((\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}-2\\\\\\(\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=2\sqrt{3}-2\\\\\\\sqrt{x}-\dfrac{1}{\sqrt{x}}=\sqrt{2\sqrt{3}-2}\)
Hence,
\(\sqrt{x}-\dfrac{1}{\sqrt{x}}=\sqrt{2\sqrt{3}-2}\)
Ques 2)
\(x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\)
on multiplying and dividing by conjugate of denominator we get:
\(x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\times \dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}+\sqrt{a-2b}}\\\\\\x=\dfrac{(\sqrt{a+2b}+\sqrt{a-2b})^2}{(\sqrt{a+2b})^2-(\sqrt{a-2b})^2}\\\\\\x=\dfrac{(\sqrt{a+2b})^2+(\sqrt{a-2b})^2+2\sqrt{a+2b}\sqrt{a-2b}}{a+2b-a+2b}\\\\\\x=\dfrac{a+2b+a-2b+2\sqrt{a+2b}\sqrt{a-2b}}{4b}\\\\\\x=\dfrac{2a+2\sqrt{a^2-4b^2}}{4b}\\\\\\x^2=(\dfrac{2a+2\sqrt{a^2-4b^2}}{4b})^2\\\\\\x^2=\dfrac{(2a+2\sqrt{a^2-4b^2})^2}{16b^2}\)
Hence, we have:
\(x^2=\dfrac{4a^2+4(a^2-4b^2)+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\x^2=\dfrac{4a^2+4a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\\\x^2=\dfrac{8a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\bx^2=\dfrac{8a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b}\\\\\\bx^2=\dfrac{8a(a+\sqrt{a^2-4b^2})-16b^2}{16b}\\\\\\bx^2=\dfrac{8a(a+\sqrt{a^2-4b^2})}{16b}-\dfrac{16b^2}{16b}\\\\\\bx^2=\dfrac{a(a+\sqrt{a^2-4b^2})}{2b}-b\\\\\\bx^2=ax-b\\\\\\i.e.\\\\\\bx^2-ax+b=0\)
