Question

rationalize denominator 5 upon 3 + root 5 minus 2 root 2​

Answer 1

Answer:

Step-by-step explanation:

5 / 3 + √5 -2√2

= 5 / 3 + √5 -2√2 × (3 + √5 + 2√2)/ 3 + √5 +2√2

= 5 (3 + √5 +2 √2) / (3 + √ 5)² -( 2 √2)²

= 5 (3 + √5 +2 √2) / 3² + √5² + 2 × 3 ×√5 - 4 ×2

= 15 + 25 √5 + 2√2 / 6 + 6√5

=15 + 25 √5 + 2√2 / 6 + 6√5 ×  6 - 6√5 / 6 - 6√5

(Rationalize again )

= 90 - 90 √5 +150 √5 -150×5 + 12√2 - 12 √10 /  6² - (6√5)²

= 12 (5 -5 √5 + √2 - √10 / 36 -180

= 5 - 5√5 + √2 - √10 /12

Answer 2

Given:

5 upon 3 + root 5 minus 2 root 2​

To find:

Rationalize denominator 5 upon 3 + root 5 minus 2 root 2​

Solution:

1) The mathematical term of the given expression is:

• $\frac{5}{3+\sqrt{5}-2\sqrt{2} }$

2) Rationalizing the denominator means that the shifting of the root from the denominator to the numerator.

3) Rationalizing:

• $\frac{5}{3+\sqrt{5}-2\sqrt{2} }$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})}{(3+\sqrt{5}-2\sqrt{2})(3+\sqrt{5}+2\sqrt{2}) }$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})}{(3+\sqrt{5})^{2} -(2\sqrt{2})^{2}}$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})}{(9+5+6\sqrt{5})-(8)}$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})}{6+6\sqrt{5}}$

The rationalized term is $\frac{5(3+\sqrt{5}+2\sqrt{2})}{6+6\sqrt{5}}$

4) We will rationalize it again as there is the root in the denominator.

• $\frac{5(3+\sqrt{5}+2\sqrt{2})}{6+6\sqrt{5}}$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})(6-6\sqrt{5})}{(6+6\sqrt{5})(6-6\sqrt{5})}$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})(6-6\sqrt{5})}{6^{2} -(6\sqrt{5})^{2} }$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})(6-6\sqrt{5})}{36 -180 }$

• $\frac{5(3+\sqrt{5}+2\sqrt{2})(6\sqrt{5}-6)}{144 }$

So the final term is $\frac{5(3+\sqrt{5}+2\sqrt{2})(6\sqrt{5}-6)}{144 }$