Nature of Roots

Here (D=(2m)^2-4(m-2)(m+2)=16), so (D\neq0) always. In exams, (D=0) is necessary for equal roots.

For a quadratic equation ax²+bx+c=0, the discriminant D=b²−4ac determines the nature of its roots. Here a=2, b=−5√2, and c=12. Thus b²=(−5√2)²=25×2=50, while 4ac=4×2×12=96. Therefore D=50−96=−46. Since the discriminant is negative, the equation has no real roots; its two roots are complex conjugates. Hence option A is correct. D=0 would indicate equal real roots, a positive discriminant would indicate two distinct real roots, and the values D=50 or D=2 are simply incorrect calculations. The irrational coefficient does not change the discriminant rule.

A real value of l can occur only when the quadratic equation has real roots, so its discriminant must satisfy D ≥ 0. Comparing l² - 2(a + 3)l + (a² + 10) with Al² + Bl + C, we have A = 1, B = -2(a + 3), and C = a² + 10. Thus D = B² - 4AC = 4(a + 3)² - 4(a² + 10) = 4[(a² + 6a + 9) - a² - 10] = 4(6a - 1). The condition 4(6a - 1) ≥ 0 gives a ≥ 1/6. Therefore option A is correct. The other choices either reverse the inequality, impose an unnecessary single value, or ignore the discriminant condition.

The governing concept is the discriminant criterion for a quadratic equation. For ax² + bx + c = 0, the roots are non-real exactly when D = b² − 4ac < 0. Here a = 1, b = −2(k − 3), and c = k² − 8k + 20. Therefore D = [−2(k − 3)]² − 4(k² − 8k + 20) = 4(k − 3)² − 4(k² − 8k + 20) = 4(2k − 11). For no real roots, 4(2k − 11) < 0. Since 4 is positive, this reduces to 2k − 11 < 0, giving k < 11/2. At k = 11/2, D = 0 and the roots are equal and real; for larger k, D > 0 and two distinct real roots occur. Thus option A is correct.

Because p ≠ 1, the coefficient p − 1 of x² is nonzero, so the equation is genuinely quadratic. Real roots require a non-negative discriminant. Here a = p − 1, b = −2(p + 1), and c = p + 3. Thus D = [−2(p + 1)]² − 4(p − 1)(p + 3). Expanding, (p + 1)² = p² + 2p + 1 and (p − 1)(p + 3) = p² + 2p − 3. Hence D = 4[(p² + 2p + 1) − (p² + 2p − 3)] = 16. This is positive for every real p. Therefore two distinct real roots exist for every allowed p, namely every p except 1. Option D is correct. Option A imposes an unsupported upper bound, and p = 1 is explicitly excluded.

Equal roots occur when the discriminant of a genuine quadratic is zero. Here a = q + 2, b = −2(q − 1), and c = q. Therefore D = [−2(q − 1)]² − 4(q + 2)q = 4(q − 1)² − 4q(q + 2). Expanding the terms gives D = 4(q² − 2q + 1 − q² − 2q) = 4(1 − 4q). Set D equal to zero: 4(1 − 4q) = 0, so 1 − 4q = 0 and q = 1/4. This value satisfies the given restriction q ≠ −2, so the coefficient of x² remains nonzero. The other options leave a nonzero discriminant and therefore do not produce equal roots. Hence option A is correct.

For a quadratic to have no real roots, its discriminant must be negative. In this equation a = 1, b = 2(t + 1), and c = 3t + 7. Hence D = [2(t + 1)]² − 4(3t + 7) = 4(t + 1)² − 4(3t + 7) = 4(t² − t − 6) = 4(t − 3)(t + 2). The constant factor 4 is positive, so D is negative precisely when (t − 3)(t + 2) is negative. A product of two linear factors is negative between its distinct zeros, −2 and 3. Therefore −2 < t < 3, making option A correct. At either endpoint D = 0 and the roots are equal and real; outside the interval D > 0 and two distinct real roots result.

The relevant principle is that a quadratic has no real roots exactly when its discriminant is negative. Here a = 1, b = −2(k − 4), and c = k² − 10k + 27. Compute D = [−2(k − 4)]² − 4(k² − 10k + 27) = 4(k − 4)² − 4(k² − 10k + 27). Since (k − 4)² = k² − 8k + 16, the expression inside the difference is k² − 8k + 16 − k² + 10k − 27 = 2k − 11. Thus D = 4(2k − 11). The condition D < 0 gives 2k − 11 < 0, so k < 11/2. Equality produces equal real roots, and larger values produce two distinct real roots. Therefore option A is correct.

The governing concept is that a quadratic has real roots when its discriminant is non-negative. Here a = p − 2, b = −2(p + 2), and c = p + 6. Compute D = [−2(p + 2)]² − 4(p − 2)(p + 6) = 4(p + 2)² − 4(p² + 4p − 12) = 4(p² + 4p + 4 − p² − 4p + 12) = 64. This calculation shows D is always positive, not 40 − 8p. Therefore, for every p ≠ 2, the equation is genuinely quadratic and has two real distinct roots. None of the supplied options states this correctly: A unnecessarily restricts p, B is opposite, C violates p ≠ 2, and D is the only option that says every p ≠ 2. Hence option D, not the supplied key A, is correct.

The governing concept is that the nature of the roots is determined by the discriminant D = B² − 4AC for a quadratic Ax² + Bx + C = 0. Comparing terms gives A = 1, B = −(a + b + 1), and C = ab + a + b. Thus D = (a + b + 1)² − 4(ab + a + b). Expanding and simplifying, D = a² + 2ab + b² + 2a + 2b + 1 − 4ab − 4a − 4b = a² − 2ab + b² + 1 − 2a − 2b = (a − b)² + 1 − 2(a + b). Therefore option A gives the correct expression. The other choices either omit terms or incorrectly claim that the discriminant is constant. Once this D is evaluated, D > 0, D = 0, or D < 0 identifies distinct real, equal, or non-real roots respectively.