A quadratic equation ax^2+bx+c=0 has two integral roots x1 and

(-b/a)^2 = (-b/a)^2 - 2(c/a) + 6
or c/a =3
This condition is satisfied by all three options
Now we want the roots to be integral, hence b^2 -4ac should be perfect square and -b+-( b^2 -4ac)^{1/2} is a factor of 2a
1. When (a,b,c) is (-1,4,-3)
b^2 -4ac= 4
-b+-( b^2 -4ac)^{1/2}= -2 and 6 both are factor of -2

2. When (a,b,c) is (1, 4, 3)
b^2 -4ac= 4
-b+-( b^2 -4ac)^{1/2}= -2 and 6 both are factor of 2

3. When (a,b,c) is (3, -10√3, 9)
b^2 -4ac=192, which is not a perfect square

IMO D

Hi jagsingh1992

You were right. Corrected now. Thanks for bringing to our notice.

Hi nick1816 & gyan one
Please help me understand how we arrived at equation
(-b/a)^2 = (-b/a)^2 - 2(c/a) + 6

C/a is product of roots and it is nowhere mentioned in the question stem
Very confused with the above equation

Posted from my mobile device

Hi Ashwani87

In the question, it is given that the square of the sum of the roots is 6 greater than the sum of the squares of the roots.
Sum of the roots = -b/a => square of the sum of the roots = (-b/a)^2
Sum of the squares of the roots = (sum of roots)^2 - 2*product of roots => Sum of the squares of the roots = (-b/a)^2 - 2c(/a)

Therefore (-b/a)^2 = (-b/a)^2 - 2(c/a) + 6

In a quadratic equation ax^2 + bx + c = 0.
The sum of the roots = -b/a
The product of the roots = c/a.
therefore c/a should be equal to 3 (satisfied in all 3)
and -b/a is the sum of both the roots, since the roots are integral III is not possible.

your logic for III option is correct but you gotta check for other two options. Sum of the roots can be integer even if the roots are rational numbers. { if a=1/2, b=3/2 then a+b =2}

A quadratic equation ax^2 + bx + c = 0 has two integral roots x1 and x2. If the square of the sum of the roots is 6 greater than the sum of the squares of the roots, which of the following could be the value of the ordered set (a, b, c)?

I. (-1, 4, -3)
II. (1, 4, 3)
III. (3, -10√3, 9)

GIVEN THAT 2x1*x2 =6,so x1*x2=3
and this is equals to c/a and satisfied by all the 3 options ,so whats next.given that all roots are integral so sum of roots(x1+x2) will be integers and is not satisfied by 3rd option ,so 1,2 will be correct.

Hi!
Thank you for the explanation but I didnt get the Sum of the squares of the roots part.
Why is it (sum of roots)^2 - 2*product of roots???
For my understanding square of the roots would be (a+b)^2 = a^2 + 2ab + b^2

Thank you in advanced!

GyanOne

Hi. The explanation for this step is as following: sum of the squares of the roots is a^2 + b^2 (not (a+b)^2= a^2+2ab+b^2, as you have stated). So we want to express a^2 + b^2, by subtracting -2ab from (a+b)^2 in order to get the desired expression. We can rewrite this portion by writing it as: a^2+b^2-2ab , and then it would be equal to a^2+b^2.

Hope that this will help.
Also, since this part seems complicating, and from this equation you only get the value of x2 = c/a, by backsolving, you can see that in each answer choice c/a = 3. This requirement is fullfiled for all answer option, so you need to focus on how to find x1 (which must be an integer). Since x1 = -b/a, the result must be an integer ( I: 4/(-1)= -4 which is an integer; II: 4/1=4 which is an iteger; III: -10√3/3 is not an integer, therefore, answer option III is not possible).

hello Bunuel Veritaskarishma please help with the basic equation formation I am perplexed.
I believe I am missing some basic understanding of the topic

m = x1
n = x2

Given: (m+n)^2 = m^2 + n^2 + 6
Rule: ax^2 + (-b/a = sum of roots)x + (c/a = product of roots) = 0

-b/a = sum of roots --> b^2 = (m+n)^2

I: b = 4, b^2 = 16
m^2 + n^2 = 10 is satisfied for values -1 and -3.

II: b = 4, b^2 = 16
m^2 + n^2 = 10 is satisfied for values 1 and 3.

III: b = (-10√3)/a = 10/√3, b^2 = 100/3
m^2 + n^2 wont be integer values.

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