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How are you getting this? I thought \((a-b)^2 = a^2 -2ab+ b^2\)
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Hope this helps.
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Sorry for buggin', but I am still curious as to why you chose to manipulate \((a-b)^2 into (a+b)^2-4ab\) when you encountered this problem? Is there some sort of method/property that comes to mind? The study guides I am using doesn't really show this, but I would love to know _________________
If my post has contributed to your learning or teaching in any way, feel free to hit the kudos button ^_^
How are you getting this? I thought \((a-b)^2 = a^2 -2ab+ b^2\)
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Hope this helps.
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Sorry for buggin', but I am still curious as to why you chose to manipulate \((a-b)^2 into (a+b)^2-4ab\) when you encountered this problem? Is there some sort of method/property that comes to mind? The study guides I am using doesn't really show this, but I would love to know
I chose this method only in this context . The question was asking for the difference of roots. . Now, we alrady know the sum and the product of the 2 roots. The formula which I have used is just to get the difference of the 2.
By the way , it might be a handy formula to remember.
How are you getting this? I thought \((a-b)^2 = a^2 -2ab+ b^2\)
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Hope this helps.
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Sorry for buggin', but I am still curious as to why you chose to manipulate \((a-b)^2 into (a+b)^2-4ab\) when you encountered this problem? Is there some sort of method/property that comes to mind? The study guides I am using doesn't really show this, but I would love to know
for a quadratic equation AX^2+BX+C = 0 SUM OF ROOTS = -B/A PRODUCT OF ROOTS = C/A let a AND b be the roots of equation then a*b = C/A a + b = -B/A
now as we have to calculate difference of roots (a-b) we can use directly the formula (a-b)^2 = (a+b)^2 - 4ab...now simply you have to plug in the values..
hope it helps _________________
When you want to succeed as bad as you want to breathe ...then you will be successfull....
Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
02 Nov 2014, 22:56
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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How are you getting this? I thought \((a-b)^2 = a^2 -2ab+ b^2\)
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Hope this helps.
\((a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2\)
Sorry for buggin', but I am still curious as to why you chose to manipulate \((a-b)^2 into (a+b)^2-4ab\) when you encountered this problem? Is there some sort of method/property that comes to mind? The study guides I am using doesn't really show this, but I would love to know
Hi Delsingh! I looked through the discussion and decided to help you if you still need help So have ever heard of Discriminant? This variable allows us to find roots of the equation without factoring. If you have a quadratic equation like a(x^2)+bx+c=0 then you can find Discriminant and roots. Formula for Discriminant is (b^2)-4ac. Formula for roots is x1=(-b+square root of Discriminant)/2a and for x2=(-b-square root of Discriminant)/2a. So you can find both roots and solve the problem. For example in our case Discriminant=25-4*(-12)*2=121. Hence x1=(-5+11)/4=1.5 and x2=(-5-11)/4=-4 Hope it is clear _________________
Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
02 Apr 2015, 11:22
2
This post received KUDOS
Expert's post
Hi All,
It looks like a number of the explanations take a more complex approach than what is needed. This question is based on FOIL-ing and Factoring rules; even though it looks a little "tough", the same rules still apply...
We're given 2X^2 + 5X = 12
We can rewrite that as....
2X^2 + 5X - 12 = 0
Now let's break this into it's two 'pieces'....
(X _ _ )(2X _ _ )
Now let's look at the '-12'....
This means that the two numbers could be.... 1 and 12 2 and 6 3 and 4
And one number is + and the other is -
Since the middle term of the Quadratic is "5X", we need to 'play around' a bit with the possibilities....
1 and 12 are too far 'apart' 2 and 6 are both even, so we won't end up with 5X (since 5 is odd)
That just leaves us with 3 and 4.... (X + 4)(2X - 3) = 0
Now we can solve the Quadratic...
X = -4, +3/2
The prompt asks for the difference in the solutions... (3/2) - (-4) = 11/2
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