given x2+1/4x2=8
find the value of x+1/2x
Answers
Step-by-step explanation:
Given:
x^{2} +\dfrac{1}{4x^{2}} =8x
2
+
4x
2
1
=8
We have to find the value of x^{3} +\dfrac{1}{8x^{3}}x
3
+
8x
3
1
= ?
Solution:
∴ x^{2} +\dfrac{1}{4x^{2}} =8x
2
+
4x
2
1
=8
⇒ x^{2} +(\dfrac{1}{2x})^2 =8x
2
+(
2x
1
)
2
=8
Using the algebraic identity:
a^{2}a
2
+ b^2b
2
= (a+b)^{2}(a+b)
2
- 2ab
(x+\dfrac{1}{2x})^2+2x.\dfrac{1}{2x}=8(x+
2x
1
)
2
+2x.
2x
1
=8
⇒ (x+\dfrac{1}{2x})^2(x+
2x
1
)
2
+ 2 = 8
⇒ (x+\dfrac{1}{2x})^2(x+
2x
1
)
2
= 8 + 2 = 10
⇒ x+\dfrac{1}{2x}x+
2x
1
= \sqrt{10}
10
∴ x^{3} +\dfrac{1}{8x^{3}}x
3
+
8x
3
1
= (x+\dfrac{1}{2x})^3-3x.\dfrac{1}{2x}(x+\dfrac{1}{2x})(x+
2x
1
)
3
−3x.
2x
1
(x+
2x
1
)
= (x+\dfrac{1}{2x})^3-3(x+\dfrac{1}{2x})(x+
2x
1
)
3
−3(x+
2x
1
)
= (\sqrt{10} )^3-3\sqrt{10}(
10
)
3
−3
10
= 10\sqrt{10}
10
- 3\sqrt{10}
10
= 7\sqrt{10}
10
∴ x^{3} +\dfrac{1}{8x^{3}}x
3
+
8x
3
1
= 7\sqrt{10}
10
Thus, the value of x^{3} +\dfrac{1}{8x^{3}}x
3
+
8x
3
1
is "equal to 7\sqrt{10}
10
".
Answer:
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