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TeamGMATIFY and chetan2u, please would you mind encoding your math illustrations...
What's the point in your solution, if i dont see what you solved?
I can tell better what it is if you encode them properly. Thanks.
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TeamGMATIFY and chetan2u, please would you mind encoding your math illustrations...
What's the point in your solution, if i dont see what you solved?
I can tell better what it is if you encode them properly. Thanks.

Edited the posts above.

Hope it helps.
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Nez
TeamGMATIFY and chetan2u, please would you mind encoding your math illustrations...
What's the point in your solution, if i dont see what you solved?
I can tell better what it is if you encode them properly. Thanks.
Hi Nez,

Sorry for the post.
Hope it helps now.
Bunuel Thanks for the editing.
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Nez
What is the value of \(9x^2 - 30xy + 25y^2\)?

1) \(3x - 5y = 20\)

2) \(x + y = 12\)


___________________________________
I need to know how you derive 3x - 5y from the question stem.
A kudos for your thought.


hi,

two ways..
1) you have three term one in x^2, one in xy and third in y^2..

also the coefficients of x^2 and y^2 are PERFECT square 9 and 25..

so \((3x)^2- 30xy+(5y)^2\)..

now \((a-b)^2=a^2-2ab+b^2\)... where \(a=3x\) and \(b=5y\)..

so \(9x^2 - 30xy + 25y^2=(3x-5y)^2\)..


2) \((3x)^2- 30xy+(5y)^2\).

=> \((3x)^2- \frac{30xy}{2} - \frac{30xy}{2} +(5y)^2\)..

=> \((3x)^2- 15xy-15xy+(5y)^2\)..

=> \((3x)^2- 3*5*x*y-3*5*x*y+(5y)^2\)..

\(3x(3x-5y)-5y(3x-5y)= (3x-5y)(3x-5y)\)


This is beauty
\((3x)^2- \frac{30xy}{2} - \frac{30xy}{2} +(5y)^2\)..

=> \((3x)^2- 15xy-15xy+(5y)^2\)..

=> \((3x)^2- 3*5*x*y-3*5*x*y+(5y)^2\)..

\(3x(3x-5y)-5y(3x-5y)= (3x-5y)(3x-5y)\)


didn't know it would not just be only easy but beautiful as well
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chetan2u
Nez
What is the value of \(9x^2 - 30xy + 25y^2\)?

1) \(3x - 5y = 20\)

2) \(x + y = 12\)


___________________________________
I need to know how you derive 3x - 5y from the question stem.
A kudos for your thought.


hi,

two ways..
1) you have three term one in x^2, one in xy and third in y^2..

also the coefficients of x^2 and y^2 are PERFECT square 9 and 25..

so \((3x)^2- 30xy+(5y)^2\)..

now \((a-b)^2=a^2-2ab+b^2\)... where \(a=3x\) and \(b=5y\)..

so \(9x^2 - 30xy + 25y^2=(3x-5y)^2\)..


2) \((3x)^2- 30xy+(5y)^2\).

=> \((3x)^2- \frac{30xy}{2} - \frac{30xy}{2} +(5y)^2\)..

=> \((3x)^2- 15xy-15xy+(5y)^2\)..

=> \((3x)^2- 3*5*x*y-3*5*x*y+(5y)^2\)..

\(3x(3x-5y)-5y(3x-5y)= (3x-5y)(3x-5y)\)


This is beauty
\((3x)^2- \frac{30xy}{2} - \frac{30xy}{2} +(5y)^2\)..

=> \((3x)^2- 15xy-15xy+(5y)^2\)..

=> \((3x)^2- 3*5*x*y-3*5*x*y+(5y)^2\)..

\(3x(3x-5y)-5y(3x-5y)= (3x-5y)(3x-5y)\)

y^2
didn't know it would not just be only easy but beautiful as well

There is a formulaic approach to solving quadratic equations as shown below:

For the quadratic equation, ax^2+bx+c=0, with x being the variable and a,b,c are constants,

Roots \(x_1\) and \(x_2\)\(= \frac {-b \pm \sqrt{b^2-4ac}}{2*a}\) and then you can write the quadratic equation as : \((x-x_1)(x-x_2)=0\)

Thus for the question above,

\(9x^2−30xy+25y^2\)---> \(a=9, b=-30y, c=25y^2\), you get,

Thus, \(x = \frac {-b \pm \sqrt{b^2-4ac}}{2*a} = \frac {30y \pm \sqrt{(30y)^2-4*9*25y^2}}{2*9} = \frac {30y \pm \sqrt{900*y^2-900*y^2}}{2*9} = \frac {30y}{2*9} = \frac {5y}{3}\)

---> You get \((x-\frac{5*y}{3})(x-\frac{5*y}{3})=0\) as the equation ---> \((3x-5y)^2 = 0\) as the perfect square.

Again, this method might seem a bit excessive for time but for some questions where factorization isnt striaghtforward, you can use this to find the roots of the quadratic equation. Alternately, in order to check whether a given qudratic equation is a perfect square is to check for the discriminant.

For a given perfect square, Discriminant, \(D= \sqrt{b^2-4ac}= 0\). In the case above as well you got D = 0, thus telling you that the given quadratic equation is a perfect square.

Hope this helps.
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Quote:
There is a formulaic approach to solving quadratic equations as shown below:

For the quadratic equation, ax^2+bx+c=0, with x being the variable and a,b,c are constants,

Roots \(x_1\) and \(x_2\)\(= \frac {-b \pm \sqrt{b^2-4ac}}{2*a}\) and then you can write the quadratic equation as : \((x-x_1)(x-x_2)=0\)

Thus for the question above,

\(9x^2−30xy+25y^2\)---> \(a=9, b=-30y, c=25y^2\), you get,

Thus, \(x = \frac {-b \pm \sqrt{b^2-4ac}}{2*a} = \frac {30y \pm \sqrt{(30y)^2-4*9*25y^2}}{2*9} = \frac {30y \pm \sqrt{900*y^2-900*y^2}}{2*9} = \frac {30y}{2*9} = \frac {5y}{3}\)

---> You get \((x-\frac{5*y}{3})(x-\frac{5*y}{3})=0\) as the equation ---> \((3x-5y)^2 = 0\) as the perfect square.

Again, this method might seem a bit excessive for time but for some questions where factorization isnt striaghtforward, you can use this to find the roots of the quadratic equation. Alternately, in order to check whether a given qudratic equation is a perfect square is to check for the discriminant.

For a given perfect square, Discriminant, \(D= \sqrt{b^2-4ac}= 0\). In the case above as well you got D = 0, thus telling you that the given quadratic equation is a perfect square.

Hope this helps.

I'm yet to come across a GMAT question requiring the quadratic formula.
Chetan has answered my question.
Kudos for your reply though
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Nez
Quote:
There is a formulaic approach to solving quadratic equations as shown below:

For the quadratic equation, ax^2+bx+c=0, with x being the variable and a,b,c are constants,

Roots \(x_1\) and \(x_2\)\(= \frac {-b \pm \sqrt{b^2-4ac}}{2*a}\) and then you can write the quadratic equation as : \((x-x_1)(x-x_2)=0\)

Thus for the question above,

\(9x^2−30xy+25y^2\)---> \(a=9, b=-30y, c=25y^2\), you get,

Thus, \(x = \frac {-b \pm \sqrt{b^2-4ac}}{2*a} = \frac {30y \pm \sqrt{(30y)^2-4*9*25y^2}}{2*9} = \frac {30y \pm \sqrt{900*y^2-900*y^2}}{2*9} = \frac {30y}{2*9} = \frac {5y}{3}\)

---> You get \((x-\frac{5*y}{3})(x-\frac{5*y}{3})=0\) as the equation ---> \((3x-5y)^2 = 0\) as the perfect square.

Again, this method might seem a bit excessive for time but for some questions where factorization isnt striaghtforward, you can use this to find the roots of the quadratic equation. Alternately, in order to check whether a given qudratic equation is a perfect square is to check for the discriminant.

For a given perfect square, Discriminant, \(D= \sqrt{b^2-4ac}= 0\). In the case above as well you got D = 0, thus telling you that the given quadratic equation is a perfect square.

Hope this helps.

I'm yet to come across a GMAT question requiring the quadratic formula.
Chetan has answered my question.
Kudos for your reply though

I mentioned the formula to give you the complete picture. Additionally, you can check whether a given quadratic equation is in fact a perfect square and not spend time on trying to coming up with a perfect square expression for a quadratic equation that is in fact not a perfect square.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

What is the value of 9x^2−30xy+25y^2 ?

1) 3x−5y=20

2) x+y=12


When you modify the original condition and the question, they become 9x^2 - 30xy + 25y^2=(3x-5y)^2?. Since 1) is unique, the answer is A.


 Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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What about the second statement? Why can we not square it and derive at a system of equations?
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Ekland
What is the value of \(9x^2 - 30xy + 25y^2\)?

1) \(3x - 5y = 20\)

2) \(x + y = 12\)


\((x-y)^2 = (x-y)(x-y)\)

\(9x^2 = (3x)^2\)

\(25y^2 = (5y)^2\)

\(30xy = 2*3x*5y\)

What is the value of \((3x-5y)(3x-5y)\)?

(1) Gives us what we need. SUFFICIENT.

(2) Can't determine x or y individually; INSUFFICIENT.

Answer is A.
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