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

Question

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)

A. I Only
B. II Only
C. III Only
D. I and II Only
E. I, II and III Only

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

A. I Only
B. II Only
C. III Only
D. I and II Only
E. I, II and III Only

GyanOne · Accepted Answer

(-b/a)^2 = (-b/a)^2 - 2(c/a) + 6 
=> c = 3a

All 3 equations satisfy this.

For all 3 options, b^2-4ac is a perfect square (real roots), but for the third option, the roots will not be integral. This rules the third option out. I and II remain.

Option (D).

jagsingh1992 · Answer

GyanOne  Hello GyanOne, tried solving the equation you have posted above in your solution. However not able to arrive at ac=3. Am getting c=3a instead. Can you kindly clarify? Thank you.

nick1816 · Answer

(-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

GyanOne · Answer

Hi  jagsingh1992

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

Ashwani87 · Answer

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

GyanOne · Answer

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

Priyanka1293 · Answer

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.

nick1816 · Answer

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}

PreiteeRanjan · Answer

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)

A. I Only
B. II Only
C. III Only
D. I and II Only
E. I, II and III Only

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.

jcerdae · Answer

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!

Biba14 · Answer

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).

shivimonji · Answer

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

Bambi2021 · Answer

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.

plpurva · Answer

the equation simplifies to 2 X1* X2= 6, or that the product of the roots  X1*X2= 3 . Since the product of roots also equals C/A, we just need to check which options have C/A= 3 and produce integer roots, which are only I and II.

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

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

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