Question

please give me correct answer SIMPLIFY​

Answer 1

Question:-

Simplify the following:-

(7 + 3√5)/(3 + √5) – (7 - 3√5)/(3 - √5)

Solution:-

$\sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}}$

Let us solve the parts one by one.

For Left part:-

= $\sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}}}$

By Rationalizing the denominator:-

= $\sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} \times \dfrac{3 - \sqrt{5}}{3 - \sqrt{3}}}$

= $\sf{\dfrac{(7 + 3\sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}}$

We know,

• (a + b)(a - b) = a² - b²

= $\sf{\dfrac{7(3 - \sqrt{5}) + 3\sqrt{5}(3 - \sqrt{5})}{(3)^2 - (\sqrt{5})^2}}$

= $\sf{\dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 15}{9 - 5}}$

= $\sf{\dfrac{6 + 2\sqrt{5}}{4}}$

= $\sf{\dfrac{2(3 + \sqrt{5}}{4}}$

= $\sf{\dfrac{3 + \sqrt{5}}{2}}$

For Left part:-

$\sf{\dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}}$

Rationalising the denominator,

= $\sf{\dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} \times \dfrac{3 + \sqrt{5}}{3 + \sqrt{3}}}$

= $\sf{\dfrac{(7 - 3\sqrt{5})( 3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}}$

We know,

• (a + b)(a - b) = a² - b²

= $\sf{\dfrac{7(3 + \sqrt{5}) - 3\sqrt{5}(3 + \sqrt{5})}{(3)^2 - (\sqrt{5})^2}}$

= $\sf{\dfrac{21 + 7\sqrt{5} - 9\sqrt{5} - 15}{9 - 5}}$

= $\sf{\dfrac{6 - 2\sqrt{5}}{4}}$

= $\sf{\dfrac{2(3 - \sqrt{5}}{4}}$

= $\sf{\dfrac{3 - \sqrt{5}}{2}}$

Therefore,

$\sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} = \dfrac{3 + \sqrt{5}}{2}}$

And,

$\sf{\dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = \dfrac{3 - \sqrt{5}}{2}}$

Hence,

The value of $\sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}}$ is as follows:-

= $\sf{\dfrac{3 + \sqrt{5}}{2} - \dfrac{3 - \sqrt{5}}{2}}$

= $\sf{\dfrac{3 + \sqrt{5} - (3 - \sqrt{5})}{2}}$

= $\sf{\dfrac{3 + \sqrt{5} - 3 + \sqrt{5}}{2}}$

= $\sf{\dfrac{2\sqrt{5}}{2}}$

= √5

∴ The answer is √5.

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Answer 2

Answer:

Question:-

Simplify the following:-

(7 + 3√5)/(3 + √5) – (7 - 3√5)/(3 - √5)

Solution:-

\sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}}

3+

5

7+3

5

−

3−

5

7−3

5

Let us solve the parts one by one.

For Left part:-

= \sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}}}

3+

5

7+3

5

By Rationalizing the denominator:-

= \sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} \times \dfrac{3 - \sqrt{5}}{3 - \sqrt{3}}}

3+

5

7+3

5

×

3−

3

3−

5

= \sf{\dfrac{(7 + 3\sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}}

(3+

5

)(3−

5

)

(7+3

5

)(3−

5

)

We know,

(a + b)(a - b) = a² - b²

= \sf{\dfrac{7(3 - \sqrt{5}) + 3\sqrt{5}(3 - \sqrt{5})}{(3)^2 - (\sqrt{5})^2}}

(3)

2

−(

5

)

2

7(3−

5

)+3

5

(3−

5

)

= \sf{\dfrac{21 - 7\sqrt{5} + 9\sqrt{5} - 15}{9 - 5}}

9−5

21−7

5

+9

5

−15

= \sf{\dfrac{6 + 2\sqrt{5}}{4}}

4

6+2

5

= \sf{\dfrac{2(3 + \sqrt{5}}{4}}

4

2(3+

5

= \sf{\dfrac{3 + \sqrt{5}}{2}}

2

3+

5

For Left part:-

\sf{\dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}}

3−

5

7−3

5

Rationalising the denominator,

= \sf{\dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} \times \dfrac{3 + \sqrt{5}}{3 + \sqrt{3}}}

3−

5

7−3

5

×

3+

3

3+

5

= \sf{\dfrac{(7 - 3\sqrt{5})( 3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}}

(3−

5

)(3+

5

)

(7−3

5

)(3+

5

)

We know,

(a + b)(a - b) = a² - b²

= \sf{\dfrac{7(3 + \sqrt{5}) - 3\sqrt{5}(3 + \sqrt{5})}{(3)^2 - (\sqrt{5})^2}}

(3)

2

−(

5

)

2

7(3+

5

)−3

5

(3+

5

)

= \sf{\dfrac{21 + 7\sqrt{5} - 9\sqrt{5} - 15}{9 - 5}}

9−5

21+7

5

−9

5

−15

= \sf{\dfrac{6 - 2\sqrt{5}}{4}}

4

6−2

5

= \sf{\dfrac{2(3 - \sqrt{5}}{4}}

4

2(3−

5

= \sf{\dfrac{3 - \sqrt{5}}{2}}

2

3−

5

Therefore,

\sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} = \dfrac{3 + \sqrt{5}}{2}}

3+

5

7+3

5

=

2

3+

5

And,

\sf{\dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = \dfrac{3 - \sqrt{5}}{2}}

3−

5

7−3

5

=

2

3−

5

Hence,

The value of \sf{\dfrac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \dfrac{7 - 3\sqrt{5}}{3 - \sqrt{5}}}

3+

5

7+3

5

−

3−

5

7−3

5

is as follows:-

= \sf{\dfrac{3 + \sqrt{5}}{2} - \dfrac{3 - \sqrt{5}}{2}}

2

3+

5

−

2

3−

5

= \sf{\dfrac{3 + \sqrt{5} - (3 - \sqrt{5})}{2}}

2

3+

5

−(3−

5

)

= \sf{\dfrac{3 + \sqrt{5} - 3 + \sqrt{5}}{2}}

2

3+

5

−3+

5

= \sf{\dfrac{2\sqrt{5}}{2}}

2

2

5

= √5

∴ The answer is √5.

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