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by sana.noor » Fri Jan 25, 2013 10:32 am
If r and s are the two roots of the equation x^2 + 8x +15 = 0, and r<s, what is the value of s-r?
a)-8
b) -2
c) 2
d) 7
e) 8
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by ceilidh.erickson » Fri Jan 25, 2013 10:58 am
We can factor the quadratic into (x + 5)(x + 3) = 0. This means that x = -5 or -3. In other words, -5 and -3 are the ROOTS of the equation.

We know that one of these is r, and the other is s. If r < s, then r must be -5 and s must be -3, because -5 < -3. So,
s - r = (-3) - (-5) = 2

The answer is C.
Last edited by ceilidh.erickson on Sat Jan 26, 2013 6:58 am, edited 1 time in total.
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by Brent@GMATPrepNow » Fri Jan 25, 2013 11:04 am
sana.noor wrote:If r and s are the two roots of the equation x^2 + 8x +15 = 0, and r<s, what is the value of s-r?
a)-8
b) -2
c) 2
d) 7
e) 8
I thought I'd point out that, if on test day you're pressed for time, or if you don't know how to answer this question (2 scenarios that happen a lot!), there are two answer choices we can eliminate immediately before guessing. Can you see which ones?
.
.
.
.
.
.
Answer: since r < s, we know that s-r must have a positive value.
As such, we can eliminate answer choices A and B, before guessing and moving on.

Cheers,
Brent
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by sana.noor » Sat Jan 26, 2013 12:12 am
Ceilidh.erickson the answer is C...Brent can you please explain the entire question
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by hemanthkumarmn » Sat Jan 26, 2013 3:06 am
x^2+8x+15=0
=> (x+3)(x+5)=0
=> x = -5 or -3

Given that s > r, from the above roots : s = -3 and r = -5

Thus s-r = -3 - (-5)
= -3 + 5
= 2.

Ceilidh.erickson interchanged the s and r value while substituting at the end.. Rest of it was correct.
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by sana.noor » Sat Jan 26, 2013 4:51 am
hemanthKumarmn, i used the same technique but i found that other people do this question in a different way, here is the sample

we know sum of roots = (-b/a) and product of roots = c/a

so here now a =1, b = 8 and c =15
s+r = -8/1
s*r = 15/1

(s-r)^2 = (s + r)^2 - 4(s*r) ==> (-8)^2 -(4*15) ==> 64 -60==>4

s-r = 2

option c

I have no idea about this technqiue "we know sum of roots = (-b/a) and product of roots = c/a" i geuss its from some theorem.
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by ceilidh.erickson » Sat Jan 26, 2013 6:58 am
My apologies! You're absolutely right. I've edited the post above. Just goes to show that even the "experts" go too quickly and make careless mistakes!
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by sana.noor » Sat Jan 26, 2013 7:43 am
Ceilidh Erickson its okay :)
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by hemant_rajput » Sat Jan 26, 2013 10:22 am
sana.noor wrote:hemanthKumarmn, i used the same technique but i found that other people do this question in a different way, here is the sample

we know sum of roots = (-b/a) and product of roots = c/a

so here now a =1, b = 8 and c =15
s+r = -8/1
s*r = 15/1

(s-r)^2 = (s + r)^2 - 4(s*r) ==> (-8)^2 -(4*15) ==> 64 -60==>4

s-r = 2

option c

I have no idea about this technqiue "we know sum of roots = (-b/a) and product of roots = c/a" i geuss its from some theorem.

Ok, its look some of you are not aware of this method. So here I'm going to throw some light on it.

Some time you are given the root of equation and you have to find the equation. Say roots are 2 and 4.

To form the equation we use this formula:-

x^2 -(sum of root)x + (product of root)

so a quadratic equation for root 2 and 4 will be.

x^2 -6x + 8 = 0


for more reference https://www.mathwarehouse.com/quadratic/ ... -roots.php
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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