Algebra

For real roots, the discriminant D = b^2 - 4 >= 0. This means b^2 >= 4, so b >= 2 or b <= -2. This allows for infinitely many values for b.

The given root is sqrt(3 + 1) = 2. Among the options, x^2 - x - 2 factors as (x - 2)(x + 1), so it has 2 as a root.

For a quadratic equation to have equal roots, its discriminant must be zero. Setting the discriminant of 2kx^2 + 2kx + 2 = 0 to zero gives (2k)^2 - 4(2k)(2) = 0, which simplifies to 4k(k - 4) = 0. Since k cannot be zero for the equation to remain quadratic, we find k = 4.

Let u = 1/x. Then u^2 + u = 12, so u^2 + u - 12 = 0. (u+4)(u-3) = 0. u = -4 or u = 3. Since u = 1/x, x = -1/4 or x = 1/3.

For repeated roots, the discriminant b^2 - 4ac must be 0. (3k)^2 - 4(9)(4) = 0 => 9k^2 = 144 => k^2 = 16 => k = +/- 4.

For a quadratic equation ax^2 + bx + c = 0 to have equal roots, the discriminant D = b^2 - 4ac must be zero. Here, (2(k+1))^2 - 4(1)(9) = 0, which simplifies to 4(k+1)^2 = 36, so (k+1)^2 = 9. Thus, k+1 = 3 or k+1 = -3, giving k = 2 or k = -4.

Expanding (x+5)(x-8) = x^2 - 3x - 40 = 0. This is a polynomial of degree 2, which is the definition of a quadratic equation.

For the product (x - a)(x - b) to be zero, either x - a = 0 or x - b = 0. The roots are x = a and x = b. The condition for these roots to be zero is a = 0 and b = 0.

If roots are 1 and 7, the equation is (p-1)(p-7) = 0, which is p^2 - 8p + 7 = 0.

If roots are 3 and 2, the equation is (x-3)(x-2) = 0, which expands to x^2 - 5x + 6 = 0.

A quadratic equation is of the form ax^2 + bx + c = 0. By the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n roots (counting multiplicity).

6 / (2y) = y / 12. Cross-multiply: 2y^2 = 72. y^2 = 36. y = 6 or -6.

Since |x| is always non-negative, x^2 + 15|x| + 14 = 0 involves only non-negative terms (x^2 >= 0, 15|x| >= 0, 14 > 0). The sum of these terms cannot be zero.