Question

I need answer for both Questions.
* It should be step-by-step..
* Answer only if you know or else leave..
* Best answer will be marked as Brainliest.
..Thanks for answering all..​

Answer 1

Step-by-step explanation:

hope it is helpful to you

Answer 2

14. Let,

• $\sf{a=2+\sqrt5}$

• $\sf{b=2-\sqrt5}$

Then,

• $\sf{x=\sqrt a+\sqrt b}$

• $\sf{y=\sqrt a-\sqrt b}$

We recall,

• $\sf{(p+q)^2+(p-q)^2=2(p^2+q^2)}$

So,

$\longrightarrow\sf{x^2+y^2=(\sqrt a+\sqrt b)^2+(\sqrt a-\sqrt b)^2}$

$\longrightarrow\sf{x^2+y^2=2(a+b)}$

$\longrightarrow\sf{x^2+y^2=2(2+\sqrt5+2-\sqrt5)}$

$\longrightarrow\underline{\underline{\sf{x^2+y^2=8}}}$

Hence 8 is the answer.

15. Let us recall rule of componendo and dividendo.

$\longrightarrow\sf{\dfrac{p}{q}=\dfrac{r}{s}\quad\iff\quad\dfrac{p+q}{p-q}=\dfrac{r+s}{r-s}}$

So here,

$\longrightarrow\sf{x=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}}$

or,

$\longrightarrow\sf{\dfrac{(x+1)+(x-1)}{(x+1)-(x-1)}=\dfrac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}}$

By rule of componendo and dividendo,

$\longrightarrow\sf{\dfrac{\sqrt{a+2b}}{\sqrt{a-2b}}=\dfrac{x+1}{x-1}}$

$\longrightarrow\sf{\sqrt{\dfrac{a+2b}{a-2b}}=\dfrac{x+1}{x-1}}$

$\longrightarrow\sf{\dfrac{a+2b}{a-2b}=\left(\dfrac{x+1}{x-1}\right)^2}$

$\longrightarrow\sf{\dfrac{a+2b}{a-2b}=\dfrac{x^2+1+2x}{x^2+1-2x}}$

Again by rule of componendo and dividendo,

$\longrightarrow\sf{\dfrac{a}{2b}=\dfrac{x^2+1}{2x}}$

$\longrightarrow\sf{\dfrac{a}{b}=\dfrac{x^2+1}{x}}$

$\longrightarrow\sf{ax=bx^2+b}$

$\longrightarrow\underline{\underline{\sf{bx^2-ax+b=0}}}$

Hence Proved!