saswata4s wrote:

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. Since x1,x2 are not integers, III is impossible.

Then (D) is our answer.

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