Question

if X equal to root 3 + 1 by root 3 minus 1 and Y equal to root 3 minus one by root 3 + 1 find the value of x square + xy minus y square

Answer 1

$x^{2}+xy+y^{2}=15$

Solution Part.

Given:

$x=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$ and $y=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

To find:

The value of $x^{2}+xy+y^{2}$

Step-by-step explanation:

Given, $x=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$

$\Rightarrow x=\dfrac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

• We have rationalized the denominator by multiplying its conjugate $(\sqrt{3}+1)$.
• We multiply the same number with the numerator also.

$\Rightarrow x=\dfrac{3+2\sqrt{3}+1}{3-1}$

$\Rightarrow x=\dfrac{4+2\sqrt{3}}{2}$

$\Rightarrow x=2+\sqrt{3}$

$\Rightarrow x^{2}=(2+\sqrt{3})^{2}$

• Squaring both sides of the relation.

$\Rightarrow x^{2}=4+4\sqrt{3}+3$

$\Rightarrow \boxed{x^{2}=7+4\sqrt{3}}$

and $y=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

$\Rightarrow y=\dfrac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$

• We have rationalized the denominator by multiplying its conjugate $(\sqrt{3}-1)$.
• We multiply the same number with the numerator also.

$\Rightarrow y=\dfrac{3-2\sqrt{3}+1}{3-1}$

$\Rightarrow y=\dfrac{4-2\sqrt{3}}{2}$

$\Rightarrow y=2-\sqrt{3}$

$\Rightarrow y^{2}=(2-\sqrt{3})^{2}$

• Squaring both sides of the relation.

$\Rightarrow y^{2}=4-4\sqrt{3}+3$

$\Rightarrow \boxed{y^{2}=7-4\sqrt{3}}$

Now, $xy=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

$\Rightarrow \boxed{xy=1}$

So, $x^{2}+xy+y^{2}$

$=7+\sqrt{3}+1+7-\sqrt{3}$

$=15$

Answer 2

Answer:

x^{2}+xy+y^{2}=15x2+xy+y2=15

Solution Part.

Given:

x=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}x=3−13+1 and y=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}y=3+13−1

To find:

The value of x^{2}+xy+y^{2}x2+xy+y2

Step-by-step explanation:

Given, x=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}x=3−13+1

\Rightarrow x=\dfrac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}⇒x=(3−1)(3+1)(3+1)(3+1)

We have rationalized the denominator by multiplying its conjugate (\sqrt{3}+1)(3+1) .We multiply the same number with the numerator also.

\Rightarrow x=\dfrac{3+2\sqrt{3}+1}{3-1}⇒x=3−13+23+1

\Rightarrow x=\dfrac{4+2\sqrt{3}}{2}⇒x=24+23

\Rightarrow x=2+\sqrt{3}⇒x=2+3

\Rightarrow x^{2}=(2+\sqrt{3})^{2}⇒x2=(2+3)2

Squaring both sides of the relation.

\Rightarrow x^{2}=4+4\sqrt{3}+3⇒x2=4+43+3

\Rightarrow \boxed{x^{2}=7+4\sqrt{3}}⇒x2=7+43

and y=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}y=3+13−1

\Rightarrow y=\dfrac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}⇒y=(3+1)(3−1)(3−1)(3−1)

We have rationalized the denominator by multiplying its conjugate (\sqrt{3}-1)(3−1) .We multiply the same number with the numerator also.

\Rightarrow y=\dfrac{3-2\sqrt{3}+1}{3-1}⇒y=3−13−23+1

\Rightarrow y=\dfrac{4-2\sqrt{3}}{2}⇒y=24−23

\Rightarrow y=2-\sqrt{3}⇒y