Question

1. x=1/(root3-root2). find rootx-(1/rootx) 2. if x=[root(a+2b)+root(a-2b)]/[root(a+2b)-root(a-2b]. show that bx^2-ax+b=0

Answer 1

1. x = 1/ ( $\sqrt{3} - \sqrt{2)}$ = $\sqrt{3}+ \sqrt{2}$;
 ( $\sqrt{x} -1/ \sqrt{x} )^{2} = x + 1/x - 2 =$ = 

Answer 2

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:

$x^2=\dfrac{4a^2+4(a^2-4b^2)+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\x^2=\dfrac{4a^2+4a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\\\x^2=\dfrac{8a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b^2}\\\\\\bx^2=\dfrac{8a^2-16b^2+8a\sqrt{a^2-4b^2}}{16b}\\\\\\bx^2=\dfrac{8a(a+\sqrt{a^2-4b^2})-16b^2}{16b}\\\\\\bx^2=\dfrac{8a(a+\sqrt{a^2-4b^2})}{16b}-\dfrac{16b^2}{16b}\\\\\\bx^2=\dfrac{a(a+\sqrt{a^2-4b^2})}{2b}-b\\\\\\bx^2=ax-b\\\\\\i.e.\\\\\\bx^2-ax+b=0$