if (x+1/x)=5 then find (x²+1/x²) and (x⁴+1/x⁴)
Answers
Correct question:-
If (x+1/x)=5,find (x²+1/x²) and (x⁴+1/x⁴)
Given:-
→(x+1/x=5)
To find:-
Value of :-
→(x²+1/x²)
→(x⁴+1/x⁴)
Solution:-
By squaring both the sides in the
eq. (x+1/x)=5,we get:-
We know that:-
=>(a+b)²=a²+b²+2ab
Here,we have got the value of (x²+1/x²)=23
Now,by squaring both the sides
in the equation (x²+1/x²)=23,we get:-
By using the same identity used above we get:-
Thus:-
→Value of (x²+1/x²) is 23.
→Value of (x⁴+1/x⁴) is 527.
If (x+1/x)=5,find (x²+1/x²) and (x⁴+1/x⁴)
Given:-
→
To find:-
Value of :-
→
→
By squaring both the sides in the
eq. (x+1/x)=5,we get:-
→ = 25=>x
2
+
x
2
1
+2×x×
x
1
=25
= > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 25=>x
2
+
x
2
1
+2=25
= > {x}^{2} + \frac{1}{ {x}^{2} } = 25 - 2=>x
2
+
x
2
1
=25−2
= > {x}^{2} + \frac{1}{ {x}^{2} } = 23=>x
2
+
x
2
1
=23
Here,we have got the value of (x²+1/x²)=23
Now,by squaring both the sides
in the equation (x²+1/x²)=23,we get:-
= > ( {x}^{2} + \frac{1}{ {x}^{2} })^{2} = (23) ^{2}=>(x
2
+
x
2
1
)
2
=(23)
2
By using the same identity used above we get:-
= > {x}^{4} + \frac{1}{ {x}^{4} } + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } = 529=>x
4
+
x
4
1
+2×x
2
×
x
2
1
=529
= > {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 529=>x
4
+
x
4
1
+2=529
= > {x}^{4} + \frac{1}{ {x}^{4} } = 529 - 2=>x
4
+
x
4
1
=529−2
= > {x}^{4} + \frac{1}{ {x}^{4} } = 527=>x
4
+
x
4
1
=527
Thus:-
→Value of (x²+1/x²) is 23.
→Value of (x⁴+1/x⁴) is 527.