I learnt this in 5th
3 x − 2 x + 1 3 x + 2 x + 1 = 5
Find the value of x satisfying the equation above.
Clarification: For this question, we are working over the complex numbers.
The answer is -1.5.
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5 solutions
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I learnt componendo and dividendo in class 5 th.. :P
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did you also learnt complex no.s at same time!!!!!
I got quadratic equation after rationalizing and removing root sign.And I got answers as 1 and -1.5.For 1 our equation has zero in denominator.
How is d=1?
i learnt in class 7
This should be easier:
3 x − 2 x + 1 3 x + 2 x + 1 = 5 ⇒ 3 x + 2 x + 1 = 5 ( 3 x − 2 x + 1 )
⇒ 6 2 x + 1 = 4 3 x ⇒ 3 2 x + 1 = 2 3 x ⇒ 9 ( 2 x + 1 ) = 4 ( 3 x )
⇒ 1 8 x + 9 = 1 2 x ⇒ 6 x = − 9 ⇒ x = − 2 3 = − 1 . 5
I did it the same way sir ! :)
My answer was the same! The only difference was that I wrote -3/2as my answer and the computer didn't accept it😂
same here,
Same here. :D
But x=-1.5 cannot satisfy this equation because you cannot have negative values inside a square root. Hence this equation has no real solution.
pp
It can by defining − 1 = i , the imaginary number. A number with both real and imaginary parts is a complex number.
also note that it is not required for numbers to be real (not specified in the question)
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But isn't x =-1.5 a real solution?
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i did not speak about solutions but instead i stated the fact that just because factors are imaginary doesn't mean that their product should follow ..... For example, you could factor x 2 + 1 as (x + i)(x-i) { which is imaginary } .... that doesn't mean it is totally unreal.
we should be a44ble to multiply [sqrt(3x) + sqrt(2x + 1)]/[sqrt(3x - sqrt(2x + 1)] by [sqrt(3x)+ sqrt(2x+1)]/[sqer(3x) + sqrt(2x+1)] =[(sqrt(3x)+sqrt(2x+1)]^2/[(3x -2x -1)]=5. Then 5(x-1) = 3x + 2sqrt(3x)sqrt(2x+1)+ 2x+1,or -6 =2sqrt(3x(2x+1), and 36 = 4(3x)(2x+1), 3 = 2x^2 + x, and 2x^2 +x -3 =0, 2x = -1 +/ sqrt(1 + 24),,2x=-6,4 Ed Gray
did you remember... Componendo and Dividendo???