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saswata4s
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A quadratic equation ax^2 + bx + c = 0 has two integral roots x1 and 05 Mar 2018, 00:09
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65% (hard)

Question Stats:

45% (01:19) correct 55% (02:43) wrong based on 20 sessions

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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



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D
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A quadratic equation ax^2 + bx + c = 0 has two integral roots x1 and Updated on: 05 Mar 2018, 07:45
Originally posted by DavidTutorexamPAL on 05 Mar 2018, 02:52.
Last edited by DavidTutorexamPAL on 05 Mar 2018, 07:45, edited 1 time in total.
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saswata4s
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
Show more

We'll translate our question stem into algebra to see what we need to do.
This is a Precise approach.

We're told that \((x1+x2)^2 = (x1)^2 + (x2)^2 + 6.\)
This simplifies into \((x1)^2 + 2(x1*x2) + (x2)^2 = (x1)^2 + (x2)^2 + 6.\)
Canceling out the factors on both sides give \(x1*x2 = 3\)

Using the equation x1,x2 = \(\frac{{-b \pm \sqrt{b^2-4ac}}}{2a}\),
I. is \(\frac{{-4 \pm \sqrt{16-12}}}{-2}\) which gives x1 = 3, x2 = 1 and x1*x2 = 3 as required.
I. is \(\frac{{-4 \pm \sqrt{16-12}}}{2}\) which gives x1 = -3, x2 = -1 and x1*x2 = 3 as required.
III. is \(\frac{{10\sqrt{3} \pm \sqrt{100*3-36*3}}}{6}\) which gives x1 = √3/3, x2 = 3√3. Since x1,x2 are not integers, III is impossible.

Then (D) is our answer.
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A quadratic equation ax^2 + bx + c = 0 has two integral roots x1 and 05 Mar 2018, 07:30
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DavidTutorexamPAL
saswata4s
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

We'll translate our question stem into algebra to see what we need to do.
This is a Precise approach.

We're told that \((x1+x2)^2 = (x1)^2 + (x2)^2 + 6.\)
This simplifies into \((x1)^2 + 2(x1*x2) + (x2)^2 = (x1)^2 + (x2)^2 + 6.\)
Canceling out the factors on both sides give \(x1*x2 = 3\)

Using the equation x1,x2 = \(\frac{{-b \pm \sqrt{b^2-4ac}}}{2a}\),
I. is \(\frac{{-4 \pm \sqrt{16-12}}}{-2}\) which gives x1 = 3, x2 = 1 and x1*x2 = 3 as required.
I. is \(\frac{{-4 \pm \sqrt{16-12}}}{2}\) which gives x1 = -3, x2 = -1 and x1*x2 = 3 as required.
III. is \(\frac{{10\sqrt{3} \pm \sqrt{100*3-36*3}}}{6}\) which gives x1 = √3/3, x2 = 3√3 and x1*x2 = 3 as required.

Then (E) is our answer.
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Hi Exampal
In question stem , it is telling "A quadratic equation ax^2 + bx + c = 0 has two integral roots x1 and x2." , but in case III roots are not integer .
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A quadratic equation ax^2 + bx + c = 0 has two integral roots x1 and 05 Mar 2018, 07:44
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sobby

Hi Exampal
In question stem , it is telling "A quadratic equation ax^2 + bx + c = 0 has two integral roots x1 and x2." , but in case III roots are not integer .
Show more

You're right, my mistake.
Fixed in the original post.

Point of fact, since in quadratic equations x1+x2 = -b/a and since in III b/a is irrational we could have eliminated it without even checking.



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