t & w q11

lev888

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m = (-b ± √(b^2 - 4ac)) / 2a

{ - ( 4xy + 8 y^2 ) +_ √ [ ( 4xy + 8 y^2 ) ^2 - 4* - (xy + 4y^2) * - 8xy ] } / ( 2 * (-xy - 4y^2 ) )
This is M.
What do you need to find?
And do you remember this? The number of questions Mini answers in x min: x/M

The result should be simplified, if possible. Not sure how to eliminate one of the 2 roots. Maybe we can prove that one of them is negative.
If you don't get the right answer it's likely you made a mistake - I didn't check your calculations (please don't ask me to do it).
 

Saumyojit

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This is M.
What do you need to find?
And do you remember this? The number of questions Mini answers in x min: x/M
yes.
x/M

If you don't get the right answer it's likely you made a mistake - I didn't check your calculations
{ - ( 8xy + 8 y^2 ) +_ √ [ ( 8xy + 8 y^2 ) ^2 - 4* - (xy + 4y^2) * - 8xy ] } / ( 2 * (-xy - 4y^2 ) )
I checked calculation once again and found it will be 8xy.


m= { - ( 8xy + 8 y^2 ) +_ √ [ ( 8xy + 8 y^2 ) ^2 - 4* - (xy + 4y^2) * - 8xy ] } / ( 2 * (-xy - 4y^2 ) )


x/m = x / ( { - ( 8xy + 8 y^2 ) +_ √ [ ( 8xy + 8 y^2 ) ^2 - 4* - (xy + 4y^2) * - 8xy ] } / ( 2 * (-xy - 4y^2 ) ) )

Do
i simplify m further?
 

Dr.Peterson

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I would simplify m before you divide x by m; do you see that you can factor 2y out of everything, and cancel it?

Then, rather than write a division, I would write x/m as x times the reciprocal.

Then, when there is a radical in the denominator, you can rationalize the denominator using the conjugate.
 

Saumyojit

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I would simplify m before you divide x by m; do you see that you can factor 2y out of everything, and cancel it?
m= { - ( 8xy + 8 y^2 ) +_ √ [ ( 8xy + 8 y^2 ) ^2 - 4* (8x^2 y^2 + 32 xy^3) ] } / ( 2 * (-xy - 4y^2 ) )

m= { - 8xy - 8 y^2 ) +_ √ [ ( 8xy + 8 y^2 ) ^2 - 4* (8x^2 y^2 + 32 xy^3) ] } / ( -2xy - 8y^2 ) )


m= { 2y ( - 4x - 4y ) +_ √ [ ( 64 x^2 y ^2 + 64 y^4 + 2 * 8xy * 8y^2 ) + ( -32x^2 y^2 - 128xy^3 ] } / ( -2xy - 8y^2 ) )

m = { 2y ( - 4x - 4y ) +_ √ [ 2y ( 32 x^2 y + 32y^3 + 64x y^2 + 2y ( -16x^2 y - 64 xy^2 ) ] } / 2y ( -x -4y )

then?
 

Dr.Peterson

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m= { - ( 8xy + 8 y^2 ) +_ √ [ ( 8xy + 8 y^2 ) ^2 - 4* (8x^2 y^2 + 32 xy^3) ] } / ( 2 * (-xy - 4y^2 ) )

m= { - 8xy - 8 y^2 ) +_ √ [ ( 8xy + 8 y^2 ) ^2 - 4* (8x^2 y^2 + 32 xy^3) ] } / ( -2xy - 8y^2 ) )


m= { 2y ( - 4x - 4y ) +_ √ [ ( 64 x^2 y ^2 + 64 y^4 + 2 * 8xy * 8y^2 ) + ( -32x^2 y^2 - 128xy^3 ] } / ( -2xy - 8y^2 ) )

m = { 2y ( - 4x - 4y ) +_ √ [ 2y ( 32 x^2 y + 32y^3 + 64x y^2 + 2y ( -16x^2 y - 64 xy^2 ) ] } / 2y ( -x -4y )

then?
You didn't factor 2y out of the radical. Are you aware that √(4y^2 ...) = 2y √(...)? Factor 4y^2 from the radicand, and take the square root of it. Then you can divide everything by it.

I'm not reading through every detail and checking, so I can't say whether what you have is right; but this is part of the work of simplifying what you have.

I'd also multiply numerator and denominator by -1 at some point, to get rid of many negatives.
 

Saumyojit

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You didn't factor 2y out of the radical. Are you aware that √(4y^2 ...) = 2y √(...)? Factor 4y^2 from the radicand, and take the square root of it. Then you can divide everything by it.
2y ( - 4x - 4y ) +_ √ (4y^2 ) * √( 16x^2 + 16 y^2 + 32xy - 8x^2 + 32xy ) / 2y ( -x - 4y )

2y { ( - 4x - 4y ) +_ √ ( 16x^2 + 16 y^2 + 32xy - 8x^2 + 32xy ) } / 2y ( - x - 4y )


( - 4x - 4y ) +_ √ ( 8x^2 + 16 y^2 + 64xy ) / ( - x - 4y )


( - 4x - 4y ) +_ √ ( (2x √2) ^2 + (2y √4) ^2 + 64xy ) / ( - x - 4y )


then?
 

Dr.Peterson

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2y ( - 4x - 4y ) +_ √ (4y^2 ) * √( 16x^2 + 16 y^2 + 32xy - 8x^2 + 32xy ) / 2y ( -x - 4y )
2y { ( - 4x - 4y ) +_ √ ( 16x^2 + 16 y^2 + 32xy - 8x^2 + 32xy ) } / 2y ( - x - 4y )
( - 4x - 4y ) +_ √ ( 8x^2 + 16 y^2 + 64xy ) / ( - x - 4y )
( - 4x - 4y ) +_ √ ( (2x √2) ^2 + (2y √4) ^2 + 64xy ) / ( - x - 4y )
then?
First, I see an error compared to my work (which is easier to see now that it is simplified); you could also see it by looking back at the choices you were given. (You have been doing that from time to time, right?? I've mentioned this before.)

So go back and look for where your 64xy came from, and correct what is probably a sign error.

Also, your last line doesn't help make anything simpler, so drop it.

Then, I recommend the other step I mentioned, multiplying by -1/-1 to eliminate the negative signs (again, you'll see that this will make your answer closer to the choices).

And then do the final thing I've already told you to do:
Then, rather than write a division, I would write x/m as x times the reciprocal.

Then, when there is a radical in the denominator, you can rationalize the denominator using the conjugate.
This will get you to something very close to the choices; but you have to decide whether either choice in the plus-or-minus is invalid.
 

Dr.Peterson

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i did not find any
Then look again, earlier. I'm not lying to you.

Have you gone through what you wrote in #69? How about the step from there to #71? You need to check every single step you write (and better as soon as you write it than later!).
 

Saumyojit

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{ ( - 4x - 4y ) +_ √ ( 16x^2 + 16 y^2 + 32xy - 8x^2 + 32xy ) } / ( - x - 4y )
Yeah , { ( - 4x - 4y ) +_ √ ( 16x^2 + 16 y^2 + 32xy - 8x^2 - 32xy ) } / ( - x - 4y )

{ ( - 4x - 4y ) +_ √ ( 16x^2 + 16 y^2 - 8x^2 ) } / ( - x - 4y )


{ ( - 4x - 4y ) +_ √ ( 8x^2 + 16 y^2 ) } / ( - x - 4y )


{ ( - 4x - 4y ) +_ √ ( 8x^2 + 16 y^2 ) } / ( - x - 4y )


{ ( - 4x - 4y ) +_ √ ( 2x (√2) ^2 + ( 2y √4 ) ^2 ) } / ( - x - 4y )


m = { ( - 4x - 4y ) +_ √ ( ( √ 4 √2 √2 ) { x^2 + y^2 * √4 ) ) } / ( - x - 4y )


m= { 2 ( - 2x - 2y ) +_ 2 √2 * √ ( x^2 + 2y^2 ) / ( - x - 4y )

Taking 2 common

m= 2 { ( - 2x - 2y ) +_ √2 * √ ( x^2 + 2y^2 ) } / ( - x - 4y )


Then? x/m
 

Dr.Peterson

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Yeah , { ( - 4x - 4y ) +_ √ ( 16x^2 + 16 y^2 + 32xy - 8x^2 - 32xy ) } / ( - x - 4y )

{ ( - 4x - 4y ) +_ √ ( 16x^2 + 16 y^2 - 8x^2 ) } / ( - x - 4y )
{ ( - 4x - 4y ) +_ √ ( 8x^2 + 16 y^2 ) } / ( - x - 4y )
{ ( - 4x - 4y ) +_ √ ( 8x^2 + 16 y^2 ) } / ( - x - 4y )
{ ( - 4x - 4y ) +_ √ ( 2x (√2) ^2 + ( 2y √4 ) ^2 ) } / ( - x - 4y )
m = { ( - 4x - 4y ) +_ √ ( ( √ 4 √2 √2 ) { x^2 + y^2 * √4 ) ) } / ( - x - 4y )
m= { 2 ( - 2x - 2y ) +_ 2 √2 * √ ( x^2 + 2y^2 ) / ( - x - 4y )

Taking 2 common
m= 2 { ( - 2x - 2y ) +_ √2 * √ ( x^2 + 2y^2 ) } / ( - x - 4y )

Then? x/m
First, look back at the choices, and you'll see that pulling √2 out of the radical doesn't help. (I wouldn't do it anyway, because it just doesn't look simpler.)

Then ...

Just keep doing what I've already said to do!
Then, I recommend the other step I mentioned, multiplying by -1/-1 to eliminate the negative signs (again, you'll see that this will make your answer closer to the choices).

And then do the final thing I've already told you to do:

... write x/m as x times the reciprocal.

Then, when there is a radical in the denominator, you can rationalize the denominator using the conjugate.
If you don't understand, ask specifically about it, rather than acting like we've never told you anything.
 

Saumyojit

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First, look back at the choices, and you'll see that pulling √2 out of the radical doesn't help.

If you don't understand, ask specifically about it, rather than acting like we've never told you anything.
Answer has come taking negative part of (b^2 - 4ac) ^1/2...

1/4[ 2 ( x+y) - ( 2 x^2 + 4 x y^2 )^1/2 ]

its finally done .
pulling √2 out of the radical doesn't help , multiplying by -1/-1 and rationalize the denominator using the conjugate are 2 most important steps .


How did you understand that pulling √2 out of the radical doesn't help?
 
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Dr.Peterson

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Q.
Mini and Vinay are quiz masters preparing for a quiz. In 'x' minutes, Mini makes 'y' questions more than Vinay. If it were possible to reduce the time needed by each to make a question by 2 mins , then in 'x' minutes Mini would make '2y' questions more than Vinay. How many questions does Mini make in 'x' minutes?

1] 1/4[ 2 ( x+y) - ( 2 x^2 + 4 y^2 )^1/2 ]
2] 1/4[ 2(x-y) - ( 2 x^2 + 4 y^2 )^1/2 ]
3] Either option 1 or 2
4] 1/4[ 2(x-y) - ( 2 x^2 - 4 y^2 )^1/2 ]
I will ask you again: Have you looked at the choices given in the problem lately? THAT is how you can tell what your answer should look like, and therefore that the √2 doesn't help you make something that looks like these answers. None of them contains √2.
 
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