find -(2)/(5) ki power 2 ×(5)/(2) ki power 2
Answers
Answer:
ANSWER:
Given:
(x - 2)/(x + 2) + (x + 2)/(x - 2) = 4
To Find:
Value of x
Solution:
We are given that,
:\implies\dfrac{x-2}{x+2}+\dfrac{x+2}{x-2}=4:⟹
x+2
x−2
+
x−2
x+2
=4
Now we'll take LCM,
:\implies\dfrac{(x-2)(x-2)+(x+2)(x+2)}{(x+2)(x-2)}=4:⟹
(x+2)(x−2)
(x−2)(x−2)+(x+2)(x+2)
=4
So,
:\implies\dfrac{(x-2)^2+(x+2)^2}{(x+2)(x-2)}=4:⟹
(x+2)(x−2)
(x−2)
2
+(x+2)
2
=4
We know that,
:\hookrightarrow(a\pm b)^2=a^2\pm2ab+b^2:↪(a±b)
2
=a
2
±2ab+b
2
And,
:\hookrightarrow(a+b)(a-b)=a^2-b^2:↪(a+b)(a−b)=a
2
−b
2
So,
:\implies\dfrac{(x-2)^2+(x+2)^2}{(x+2)(x-2)}=4:⟹
(x+2)(x−2)
(x−2)
2
+(x+2)
2
=4
:\implies\dfrac{(x^2-4x+4)+(x^2+4x+4)}{(x^2-4)}=4:⟹
(x
2
−4)
(x
2
−4x+4)+(x
2
+4x+4)
=4
Opening brackets,
:\implies\dfrac{x^2-4x+4+x^2+4x+4}{x^2-4}=4:⟹
x
2
−4
x
2
−4x+4+x
2
+4x+4
=4
:\implies\dfrac{2x^2+8}{x^2-4}=4:⟹
x
2
−4
2x
2
+8
=4
Transposing x² - 4 to RHS,
:\implies2\!\!\!/\:(x^2+4)=4\!\!\!/^2\:(x^2-4):⟹2/(x
2
+4)=4/
2
(x
2
−4)
So,
:\implies x^2+4=2(x^2-4):⟹x
2
+4=2(x
2
−4)
Opening the brackets,
:\implies x^2+4=2x^2-8:⟹x
2
+4=2x
2
−8
Transposing LHS to RHS,
:\implies 0=2x^2-8-(x^2+4):⟹0=2x
2
−8−(x
2
+4)
So,
:\implies 0=2x^2-8-x^2-4:⟹0=2x
2
−8−x
2
−4
:\implies x^2-12=0:⟹x
2
−12=0
Transposing 12 to RHS,
:\implies x^2=12:⟹x
2
=12
:\implies x=\pm\sqrt{12}:⟹x=±
12
Therefore,
:\implies\bf x=\pm2\sqrt2:⟹x=±2
2
Hence, the value of x is ±2√2.