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/ (
`√(3) - √(2))` =
`√(3)+ √(2)`;
(
`√(x) -1/ √(x) )^(2) = x + 1/x - 2 =` =

Answer 2

Answer with explanation:

Ques 1)

`x=(1)/(√(3)-√(2))`

Now we are asked to find the value of:

`√(x)-(1)/(√(x))`

We know that:

`(√(x)-(1)/(√(x)))^2=x+(1)/(x)-2`

Also:

`x=(1)/(√(3)-√(2))` could be written as:

`x=(1)/(√(3)-√(2))* (√(3)+√(2))/(√(3)+√(2))\\\\\\x=(√(3)+√(2))/((√(3))^2-(√(2))^2)`

since, we know that:

`(a+b)(a-b)=a^2-b^2`

Hence,

`x=(√(3)+√(2))/(3-2)\\\\\\x=√(3)+√(2)`

Also,

`(1)/(x)=√(3)-√(2)`

Hence, we get:

`(√(x)-(1)/(√(x)))^2=√(3)+√(2)+√(3)-√(2)-2\\\\\\(√(x)-(1)/(√(x)))^2=2√(3)-2\\\\\\√(x)-(1)/(√(x))=\sqrt{2√(3)-2}`

Hence,

`√(x)-(1)/(√(x))=\sqrt{2√(3)-2}`

Ques 2)

`x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))`

on multiplying and dividing by conjugate of denominator we get:

`x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))* (√(a+2b)+√(a-2b))/(√(a+2b)+√(a-2b))\\\\\\x=((√(a+2b)+√(a-2b))^2)/((√(a+2b))^2-(√(a-2b))^2)\\\\\\x=((√(a+2b))^2+(√(a-2b))^2+2√(a+2b)√(a-2b))/(a+2b-a+2b)\\\\\\x=(a+2b+a-2b+2√(a+2b)√(a-2b))/(4b)\\\\\\x=(2a+2√(a^2-4b^2))/(4b)\\\\\\x^2=((2a+2√(a^2-4b^2))/(4b))^2\\\\\\x^2=((2a+2√(a^2-4b^2))^2)/(16b^2)`

Hence, we have:

`x^2=(4a^2+4(a^2-4b^2)+8a√(a^2-4b^2))/(16b^2)\\\\\\x^2=(4a^2+4a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\\\\\\\\x^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\\\\\\bx^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b)\\\\\\bx^2=(8a(a+√(a^2-4b^2))-16b^2)/(16b)\\\\\\bx^2=(8a(a+√(a^2-4b^2)))/(16b)-(16b^2)/(16b)\\\\\\bx^2=(a(a+√(a^2-4b^2)))/(2b)-b\\\\\\bx^2=ax-b\\\\\\i.e.\\\\\\bx^2-ax+b=0`