Algebra

For x^2 + px - 3 = 0, x1 + x2 = -p and x1 * x2 = -3. We know x1^2 + x2^2 = (x1 + x2)^2 - 2(x1 * x2). So 10 = (-p)^2 - 2(-3) = p^2 + 6. p^2 = 4, so p = 2 or -2.

For mx^2 + 6x + (2m-1) = 0, the product of roots is c/a = (2m-1)/m. Given product = -1, (2m-1)/m = -1 => 2m-1 = -m => 3m = 1 => m = 1/3.

Given x, y are roots of x^2 + bx + 1 = 0, we have x+y = -b and xy = 1. The expression is (x+b+y+b) / ((x+b)(y+b)) = (x+y+2b) / (xy + b(x+y) + b^2). Substituting values: (-b + 2b) / (1 + b(-b) + b^2) = b / (1 - b^2 + b^2) = b/1 = b.

If x is a common root, x^5 + ax + 1 = 0 and x^6 + ax^2 + 1 = 0. Multiply the first by x: x^6 + ax^2 + x = 0. Subtract the second equation: (x^6 + ax^2 + x) - (x^6 + ax^2 + 1) = 0 => x - 1 = 0 => x = 1. Substitute x=1 into the first equation: 1^5 + a(1) + 1 = 0 => a + 2 = 0 => a = -2.

For equal roots, the discriminant D = b^2 - 4ac must be 0. Here, (a+3)^2 - 4(2)(8) = 0, so (a+3)^2 - 64 = 0. Thus (a+3)^2 = 64, meaning a+3 = 8 or a+3 = -8. This gives a = 5 or a = -11.

Let u = |x-3|. u^2 + u - 6 = 0 => (u+3)(u-2) = 0. u = 2. |x-3| = 2 => x-3 = 2 or x-3 = -2. x = 5 or x = 1. Sum = 6.

Expanding the equation gives 3x^2 - 2(a+b+c)x + (ab+bc+ca) = 0. For equal roots, the discriminant D = 0. So, 4(a+b+c)^2 - 4*3*(ab+bc+ca) = 0, which simplifies to a^2+b^2+c^2 - ab - bc - ca = 0. This implies a=b=c, meaning a^2+b^2+c^2 = ab+bc+ca.

The root is 1/(1+sqrt(3)). Rationalizing the denominator gives (sqrt(3)-1)/(3-1) = (sqrt(3)-1)/2. If x = (sqrt(3)-1)/2, then 2x + 1 = sqrt(3). Squaring both sides: (2x+1)^2 = 3, so 4x^2 + 4x + 1 = 3, which simplifies to 4x^2 + 4x - 2 = 0, or 2x^2 + 2x - 1 = 0.

For x^2 + 3x - 4 = 0, sum of roots alpha + beta = -3, product alpha * beta = -4. 1/alpha + 1/beta = (alpha + beta) / (alpha * beta) = -3 / -4 = 3/4.

Let roots of 2x^2 + 3x + 1 = 0 be a, b. a+b = -3/2, ab = 1/2. We want an equation with roots a^2, b^2. Sum = a^2+b^2 = (a+b)^2 - 2ab = 9/4 - 1 = 5/4. Product = a^2b^2 = (ab)^2 = 1/4. Equation: x^2 - (5/4)x + 1/4 = 0, or 4x^2 - 5x + 1 = 0.

If p, q, r are in AP, then 2q = p + r. Roots sum alpha + beta = -q/p = 4, so q = -4p. Then 2(-4p) = p + r => r = -9p. Product alpha * beta = r/p = -9p/p = -9.

For x^2 + px + c = 0, roots are 2, -2, so c = product = -4. For x^2 + bx + q = 0, roots are -1, -2, so b = -(sum) = 3. The equation x^2 + bx + c = 0 becomes x^2 + 3x - 4 = 0. Factoring gives (x+4)(x-1) = 0, so roots are -4, 1.

Given a^2 - 3a + 1 = 0, we have a + 1/a = 3. Dividing the numerator and denominator of the expression by a^3 gives 1 / (a^3 + 1/a^3). Using the identity a^3 + 1/a^3 = (a + 1/a)^3 - 3(a + 1/a), we get 3^3 - 3(3) = 27 - 9 = 18, so the value is 1/18.

Sum of roots = -b/a = -(2a+3)/(a+1) = -1. 2a+3 = a+1 -> a = -2. Product of roots = c/a = (3a+4)/(a+1). Substitute a = -2: (3(-2)+4)/(-2+1) = -2/-1 = 2.