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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
04 Aug 2013, 10:57

mau5 wrote:

You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.

Note that (a-b)^2 = (a+b)^2-4ab

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
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
04 Aug 2013, 11:06

Expert's post

DelSingh wrote:

mau5 wrote:

You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.

Note that (a-b)^2 = (a+b)^2-4ab

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.

Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
04 Aug 2013, 11:07

2

This post received KUDOS

DelSingh wrote:

mau5 wrote:

You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.

Note that (a-b)^2 = (a+b)^2-4ab

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
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