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Using componendo and dividendo solve for x: (sqrt(2x + 2) + sqrt(2x - 1))/(sqrt(2x + 2) - sqrt(2x - 1)) = 3 - Mathematics

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Using componendo and dividendo solve for x:

`(sqrt(2x + 2) + sqrt(2x - 1))/(sqrt(2x + 2) - sqrt(2x - 1))` = 3

Using componendo and dividendo rule, solve for ‘x’

`(sqrt(2x + 2) + sqrt(2x - 1))/(sqrt(2x + 2) - sqrt(2x - 1))` = 3

Sum
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Solution

`(sqrt(2x + 2) + sqrt(2x - 1))/(sqrt(2x + 2) - sqrt(2x - 1)) = 3/1`

Using componendo and dividendo,

`((sqrt(2x + 2) + sqrt(2x - 1)) + (sqrt(2x + 2) - sqrt(2x - 1)))/((sqrt(2x + 2) + sqrt(2x - 1)) - (sqrt(2x + 2) - sqrt(2x - 1))) = (3 + 1)/(3 - 1)`

`(2sqrt(2x + 2))/(2sqrt(2x - 1)) = 4/2`

`sqrt(2x + 2)/(sqrt(2x - 1)` = 2

Square both sides,

`(sqrt(2x + 2)/sqrt(2x - 1))^2 = 2^2`

`(2x + 2)/(2x - 1)` = 4

2x + 2 = 4(2x − 1)

2x + 2 = 8x – 4

2x – 8x = – 2 – 4

– 6x =  – 6

x = 1

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Chapter 7: Ratio and proportion - Chapter Test [Page 143]

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Nootan Mathematics [English] Class 10 ICSE
Chapter 7 Ratio and proportion
Chapter Test | Q 1. | Page 143
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