Question

if p=3-√5/3+√5 and q=3+√5/3-√5, find the value of p^2+q^2​

Answer 1

Answer:

47

Step-by-step explanation:

As per the provided information in the given question,

● $\longmapsto\rm {p = \dfrac{3-\sqrt{5}}{3+\sqrt{5}} }$

● $\longmapsto\rm {q = \dfrac{3+\sqrt{5}}{3-\sqrt{5}} }$

We are asked to calculate the value of p² + q².

We know that,

→ (p + q)² = p² + q² + 2pq

→ (p + q)² - 2pq = p² + q²

Finding the value of (p + q)² :

$\longmapsto\rm {(p + q)^2 = { \Bigg [ \dfrac{3-\sqrt{5}}{3+\sqrt{5}} + \dfrac{3+\sqrt{5}}{3-\sqrt{5}}\Bigg ] }^2}$

$\longmapsto\rm {(p + q)^2= { \Bigg [ \dfrac{(3-\sqrt{5})^2 + (3 + \sqrt{5})^2}{(3+\sqrt{5})(3- \sqrt{5}) } \Bigg ] }^2}$

$\longmapsto\rm { (p + q)^2= { \Bigg [ \dfrac{(9 + 5- 6\sqrt{5}) + (9 + 5+ 6\sqrt{5}) }{(3)^2 - (\sqrt{5})^2 } \Bigg ] }^2}$

$\longmapsto\rm { (p + q)^2= { \Bigg [ \dfrac{9 + 5\cancel{- 6\sqrt{5}} + 9 + 5\cancel{+ 6\sqrt{5}} }{9 -5 } \Bigg ] }^2}$

$\longmapsto\rm { (p + q)^2= { \Bigg [ \dfrac{9 + 5+ 9 + 5}{4 } \Bigg ] }^2}$

$\longmapsto\rm { (p + q)^2= { \Bigg [ \dfrac{14+ 14}{4 } \Bigg ] }^2}$

$\longmapsto\rm { (p + q)^2= { \Bigg [ \dfrac{28}{4 } \Bigg ] }^2}$

$\longmapsto\rm { (p + q)^2= {(7)}^2}$

$\longmapsto\bf { (p + q)^2= 49 }$

Finding the value of 2pq :

$\longmapsto\rm { 2pq= 2 \Bigg ( \dfrac{\cancel{3-\sqrt{5}}}{\cancel{3+\sqrt{5}}} \times \dfrac{\cancel{3+\sqrt{5}}}{\cancel{3-\sqrt{5}} } \Bigg )}$

$\longmapsto\rm { 2pq = 2(1)}$

$\longmapsto\bf { 2pq = 2}$

Finding the value of p² + q² :

→ (p + q)² - 2pq = p² + q²

→ 49 - 2 = p² + q²

→ 47 = p² + q²

Therefore, the required answer is 47.

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