Advertisements
Advertisements
प्रश्न
Using componendo and dividendo, find the value of x:
`(sqrt(3x + 4) + sqrt(3x - 5))/(sqrt(3x + 4) - sqrt(3x - 5)) = 9`
उत्तर
`(sqrt(3x + 4) + sqrt(3x - 5))/(sqrt(3x + 4) - sqrt(3x - 5)) = 9/1`
Applying componendo and dividend, we have
`(sqrt(3x + 4) + sqrt(3x - 5) + sqrt(3x + 4) - sqrt(3x - 5))/(sqrt(3x + 4) + sqrt(3x - 5) - sqrt(3x + 4) + sqrt(3x - 5)) = (9 + 1)/(9 - 1)`
`=> (2sqrt(3x + 4))/(2sqrt(3x - 5)) = 5/4`
Squaring both sides, we have
`(3x + 4)/(3x - 5) = 25/16`
`=>` 16(3x + 4) = 25(3x – 5)
`=>` 48x – 64 = 75x – 125
`=>` 75x – 48x = 64 + 125
`=>` 27x = 189
`=> x = 189/27`
`=>` x = 7
APPEARS IN
संबंधित प्रश्न
If `(a - 2b - 3c + 4d)/(a + 2b - 3c - 4d) = (a - 2b + 3c - 4d)/(a + 2b + 3c + 4d)`, show that: 2ad = 3bc.
Given x = `(sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) + sqrt(a^2 - b^2))`
Use componendo and dividendo to prove that b^2 = (2a^2x)/(x^2 + 1)
If `x = (sqrt(a + 1) + sqrt(a - 1))/(sqrt(a + 1) - sqrt(a - 1))`, using properties of proportion show that: x2 – 2ax + 1 = 0.
If p, q, r ands are In continued proportion, then prove that (p3+q3+r3) ( q3+r3+s3) : : P : s
Find x, if `16((a - x)/(a + x))^3 = (a + x)/(a - x)`.
If a : b : : c : d, prove that (la + mb) : (lc + mb) :: (la – mb) : (lc – mb)
If (ma + nb): b :: (mc + nd) : d, prove that a, b, c, d are in proportion.
Find x from the following equations : `(sqrt(a + x) + sqrt(a - x))/(sqrt(a + x) - sqrt(a - x)) = c/d`
Given `x = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`. Use componendo and dividendo to prove that: `b^2 = (2a^2x)/(x^2 + 1)`
Given that `(a^3 + 3ab^2)/(b^3 + 3a^2b) = (63)/(62)`. Using componendo and dividendo find a: b.
