if x^2+1/x^2=7 find value of (x+1/x), (x-1/x),(2x^2-2/x^2)
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Answered by
0
Answer:
Ifx
2
+1/x
2
−7 find the valve of x
3
+1/3
3
Given equation is x
2
+1/x
2
=7
we know that(x+1/x)
2
=x
2
+2.x.1/x+(1/x)
2
=x
2
+2+1/x
2
=(x
2
+1/x
2
)+2
=7+2
=9.
∴(x+1/x)
2
=9
⇒(x+1/x)=
(3)
x
⇒(x+1/x)=3.
Now ,
(x
3
+1/x)
3
=(x+1/x)
3
−3.x.1/x.(x+1/x)
=(x+1/x)
3
−3(x+1/x)
=(3)
3
−3(3)
=27−9
=18.
∴x
3
+1/x
3
=18.
Answered by
0
Step-by-step explanation:
Given: x² + 1/x² = 7
1.) (x+ 1/x)² Using, (a+b)² = a²+b²+2ab
= x² + 1/x² + 2×x×1/x = x² + 1/x² + 2 = 7+2 =9
=> x+1/x = 3
2.) (x-1/x)² = x² + 1/x² - 2 = 7-2 = 5
=> x- 1/x = root5
3.) 2x² - 2/x² = 2(x² - 1/x²)
Using, a²-b² = (a+b)(a-b)
= 2( x+1/x )×(x-1/x)
= 2×3×root5 = 6root5
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