if 
then the value of
is
Answers
Answer:
34
Explanation:
x = 3+ √8
x = (3+√8)(3-√8) / (3-√8)
x = (9 -8) / (3-√8)
x = 1/ (3 -√8)
Or we can write 1/x = (3-√8)
Put the value and find out
x² +(1/x²) = x² + (1/x)²
So x² + 1/ x² = (3 +√8)² + (3 -√8)²
= 9+6√8+8 + 9–6√8+8
= 34.
Explanation:
Given:-
\sf\to x = 3+\sqrt{8}→x=3+
8
To find:-
\sf\to x^2+\dfrac{1}{x^2}→x
2
+
x
2
1
Now we can write as
\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}→
x
1
=
3+
8
1
Now rationalise
\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}\times\dfrac{3-\sqrt{8}}{3-\sqrt{8}}=\dfrac{3-\sqrt{8}}{9-8} =3-\sqrt{8}→
x
1
=
3+
8
1
×
3−
8
3−
8
=
9−8
3−
8
=3−
8
we get
\sf\to\dfrac{1}{x} =3-\sqrt{8}→
x
1
=3−
8
Using this identities
\to\sf (a+b)^2= a^2+b^2+2ab→(a+b)
2
=a
2
+b
2
+2ab
Now we can write as
\sf\to\bigg(x+\dfrac{1}{x}\bigg)^2=x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}→(x+
x
1
)
2
=x
2
+
x
2
1
+2×x×
x
1
Now put the value of x and 1/x
\sf\to (3+\sqrt{8}+3-\sqrt{8})^2=x^2+\dfrac{1}{x^2}→(3+
8
+3−
8
)
2
=x
2
+
x
2
1
\sf\to (3+3)^2=x^2+\dfrac{1}{x^2} +2→(3+3)
2
=x
2
+
x
2
1
+2
\sf\to (6)^2=x^2+\dfrac{1}{x^2}→(6)
2
=x
2
+
x
2
1
\sf\to 36=x^2+\dfrac{1}{x^2} +2→36=x
2
+
x
2
1
+2
\sf\to x^2+\dfrac{1}{x^2} =36-2→x
2
+
x
2
1
=36−2
\sf\to x^2+\dfrac{1}{x^2} =34→x
2
+
x
2
1
=34
Answer:-
\sf\to x^2+\dfrac{1}{x^2}→x
2
+
x
2
1