Question

If x = 3+2√2, find the value of x2 + 1/x2​

this question is of maths plz ans. fast

Answer 1

Explanation:

x = 3+2√2

=> x²= (3+2√2)²

= ( 9+8+12√2)

= 17+12√2

=> 1/x² = 1/(17+12√2)

therefore x²+1/x² = (17+12√2) + 1/(17+12√2)

=> (17+12√2) + (17-12√2) / (289-288)

= 17+17

= 34

hope this helps you

Answer 2

$\large\bf\underline \orange{Given:-}$

• x = 3+2√2

$\large\bf\underline \orange{To \: find:-}$

• x² + 1/x²

$\huge\bf\underline \green{Solution:-}$

$: \longmapsto \sf\: {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ : \longmapsto \sf\:(3 + 2 \sqrt{2}) {}^{2} + \frac{1}{ {(3 + 2 \sqrt{2}) }^{2} } \\ \\ \bf\: \red{(a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab} \\ \\ : \longmapsto \rm\: \small {3}^{2} + ( {2 \sqrt{2} )}^{2} + 2(3)(2 \sqrt{2} ) + \small\frac{1}{ {3}^{2} + (2 \sqrt{2} ) {}^{2} + 2 \times 3 \times 2 \sqrt{2} } \\ \\ : \longmapsto \sf\small9 + 8 + 12 \sqrt{2} + \frac{1}{9 + 8 + 12 \sqrt{2} } \\ \\: \longmapsto \sf\:17 + 12 \sqrt{2} + \frac{1}{17 + 12 \sqrt{2} } \\ \\ \mapsto 17 + 12 \sqrt{2} + \frac{1}{17 + 12 \sqrt{2} } \times \frac{17 - 12 \sqrt{2} }{17 - 12 \sqrt{2} } \\ \\ : \longmapsto \bf\: \pink{ {(a + b)}(a - b) = {a}^{2} - {b}^{2}} \\ \\ : \longmapsto \sf\:17 + 12 \sqrt{2} + \frac{17 - 12 \sqrt{2} }{289 - 144 \times 2} \\ \\ : \longmapsto \sf\:17 + 12 \sqrt{2} + \frac{17 - 12\sqrt{2} }{289 - 288} \\ \\ : \longmapsto \sf\:17 + \cancel{ 12 \sqrt{2}} + 17 - \cancel{ 12 \sqrt{2}} \\ \\ : \longmapsto \bf \blue{\:34}$