Algebra

For equal roots, the discriminant D = b^2 - 4ac must be 0. Here, (-2k)^2 - 4(1)(5k) = 4k^2 - 20k = 0. Factoring gives 4k(k - 5) = 0, so k = 0 or k = 5.

Ramesh read the constant term wrong, so the coefficient of x (sum of roots) is correct. Sum = 8 + 2 = 10, so the equation is x^2 - 10x + c = 0. Mahesh read the coefficient of x wrong, so the constant term (product of roots) is correct. Product = -11 * 1 = -11. The equation is x^2 - 10x - 11 = 0. Roots are (x-11)(x+1) = 0, so 11 and -1.

For the roots to be common, the coefficients must be proportional. Comparing the two equations, we find the ratio of coefficients leads to 2r - p = 0.

If x is a common root, x^2 - 4x + 1 = 0. From the second equation, p(x^2 - 4x + 1) + (4px - px - p^2x - 3x + 2p^2 - p) = 0. This simplifies to x(3p - p^2 - 3) + 2p^2 - p = 0. Substituting x^2 = 4x - 1, we find p=1 satisfies the condition.

If the equations have a common root alpha, then alpha^2 + a*alpha + b = 0 and alpha^2 + b*alpha + a = 0. Subtracting the two gives (a-b)*alpha + (b-a) = 0, so (a-b)(alpha - 1) = 0. If a != b, then alpha = 1. Substituting alpha = 1 into the first equation gives 1 + a + b = 0, so a + b = -1.

For the equations to have integer roots, the discriminants must be perfect squares. Let D1 = m^2 + 4n = k1^2 and D2 = m^2 - 4n = k2^2. Adding these gives 2m^2 = k1^2 + k2^2, and subtracting gives 8n = k1^2 - k2^2. Testing small integer values for m and n shows that n must be a multiple of 6.

If the equations have a common root, the roots of 5x^2 + 12x + 13 = 0 are complex (D = 144 - 260 < 0). If a, b, c are sides of a triangle, they must be real. This implies the equations are identical, so a/5 = b/12 = c/13. Sides are 5k, 12k, 13k. Since 5^2 + 12^2 = 13^2, it is a right-angled triangle.

Since 2 is a root of x^2 + bx + 12 = 0, 4 + 2b + 12 = 0, so b = -8. The second equation is x^2 - 8x + q = 0. For equal roots, the discriminant b^2 - 4ac must be 0. Thus, (-8)^2 - 4(1)(q) = 0, which means 64 = 4q, so q = 16.

The roots -2 and -15 correspond to the equation x^2 + 17x + 30 = 0 (since sum = -17, product = 30). The original equation had 13 instead of 17, so it is x^2 + 13x + 30 = 0. Factoring gives (x + 10)(x + 3) = 0, so the roots are -3 and -10.

The roots -15 and -4 imply the equation x^2 + 19x + 60 = 0. The student read the coefficient of x wrongly as 19 instead of 16. The correct equation is x^2 + 16x + 60 = 0. Factoring gives (x+6)(x+10) = 0, so the roots are -6 and -10.

Substitute x = -3 into both equations: y = (-3)^2 + 4(-3) + 3 = 9 - 12 + 3 = 0. y = 2(-3) + 6 = -6 + 6 = 0. Since both match, (-3, 0) is a solution.

If they have a common root x, then x^2+3x+2 = 0 => (x+1)(x+2)=0, so x = -1 or x = -2. If x = -1, 1 - k + 6 = 0 => k = 7. If x = -2, 4 - 2k + 6 = 0 => 2k = 10 => k = 5. Values are 5, 7.