Advertisements
Advertisements
Question
Find x from the following equations : `(sqrt(2 - x) + sqrt(2 + x))/(sqrt(2 - x) - sqrt(2 + x)` = 3
Solution
`(sqrt(2 - x) + sqrt(2 + x))/(sqrt(2 - x) - sqrt(2 + x)` = 3
Applying componendo and dividendo,
`(sqrt(2 - x) + sqrt(2 + x) + sqrt(2 - x) - sqrt(2 + x))/(sqrt(2 - x) + sqrt(2 + x) - sqrt(2 - x) + sqrt(2 + x)) = (3 + 1)/(3 - 1)`
⇒ `(2sqrt(2 - x))/(2sqrt(2 + x)) = (4)/(2)`
⇒ `sqrt(2 - x)/(sqrt(2 + x)) = (2)/(1)`
Squaring both sides
`(2 - x)/(2 + x) = (4)/(1)`
⇒ 8 + 4x = 2 - x
4x + x = 2
⇒ 5x = –6
∴ x = `(-6)/(5)`.
APPEARS IN
RELATED QUESTIONS
If `x = (6ab)/(a + b)`, find the value of `(x + 3a)/(x - 3a) + (x + 3b)/(x - 3b)`.
If 7x – 15y = 4x + y, find the value of x : y. Hence, use componendo and dividend to find the values of:
`(3x^2 + 2y^2)/(3x^2 - 2y^2)`
If `(x^2 + y^2)/(x^2 - y^2) = 2 1/8`, find: `(x^3 + y^3)/(x^3 - y^3)`
If a : b : : c : d, then prove that
`(4"a" + 9"b")/(4"c" + 9"d") = (4"a" - 9"b")/(4"c" - 9"d")`
If a : b = c : d , then prove that `("a"^2 + "ab" +
"b"^2)/("a"^2 - "ab" + "b"^2) = ("c"^2 + "cd"+ "d"^2)/("c"^2 - "cd" + "d"^2)`
Given that `(a^3 + 3ab^2)/(b^2 + 3a^2b) = (63)/(62)`.
Using Componendo and Dividendo find a : b.
If `(3x + 5y)/(3x - 5y) = (7)/(3)`, find x : y.
Solve for x : `(1 - px)/(1 + px) = sqrt((1 + qx)/(1 - qx)`
If a : b : : c : d, prove that (2a + 3b)(2c – 3d) = (2a – 3b)(2c + 3d)
If x = `(8ab)/"a + b"` find the value of `(x + 4a)/(x - 4a) + (x + 4b)/(x - 4b)`