Answer :
1. x = 1/ ( [tex] \sqrt{3} - \sqrt{2)} [/tex] = [tex] \sqrt{3}+ \sqrt{2} [/tex];
( [tex] \sqrt{x} -1/ \sqrt{x} )^{2} = x + 1/x - 2 =[/tex] =
( [tex] \sqrt{x} -1/ \sqrt{x} )^{2} = x + 1/x - 2 =[/tex] =
Answer with explanation:
Ques 1)
[tex]x=\dfrac{1}{\sqrt{3}-\sqrt{2}}[/tex]
Now we are asked to find the value of:
[tex]\sqrt{x}-\dfrac{1}{\sqrt{x}}[/tex]
We know that:
[tex](\sqrt{x}-\dfrac{1}{\sqrt{x}})^2=x+\dfrac{1}{x}-2[/tex]
Also:
[tex]x=\dfrac{1}{\sqrt{3}-\sqrt{2}}[/tex] could be written as:
[tex]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}[/tex]
since, we know that:
[tex](a+b)(a-b)=a^2-b^2[/tex]
Hence,
[tex]x=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}\\\\\\x=\sqrt{3}+\sqrt{2}[/tex]
Also,
[tex]\dfrac{1}{x}=\sqrt{3}-\sqrt{2}[/tex]
Hence, we get:
[tex](\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}[/tex]
Hence,
[tex]\sqrt{x}-\dfrac{1}{\sqrt{x}}=\sqrt{2\sqrt{3}-2}[/tex]
Ques 2)
[tex]x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}[/tex]
on multiplying and dividing by conjugate of denominator we get:
[tex]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}[/tex]
Hence, we have:
[tex]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[/tex]
