Quadratic equations practice · Mixed difficulty

Option D (None of these) is correct because the roots provided are not correct.

Option B (16) is correct because 16 exceeds its positive square root (4) by 12 (16 = 4 + 12).

The correct answer is C (2) because when k=2, the product of roots is 7 and the roots are real.

Option A (1) is correct because the equation has only one real solution, x = 7/3, after excluding x = ±2.

Option A (4) is correct because substituting x = -1 into the polynomial and setting it to zero allows us to solve for p.

Option D is correct because the set of possible values of 'a' for which both the sum and product of the roots are less than 1 is the interval (1, 5/2).

Option A ($ac{x^2} - bx + 1 = 0$) is correct because it is the transformed equation obtained by replacing $x$ with $\frac{{1 - bx}}{{ax}}$ in the original equation.

Option A (0) is correct because the range of the left side is less than or equal to 2, while the range of the right side is greater than or equal to 2, and they never equal each other except at a single point that doesn't satisfy the equation.

Option A ($(3, \infty)$) is correct because the eccentricity of an ellipse is less than 1 and the eccentricity of a hyperbola is greater than 1, leading to the condition a > 3.

Option A (2) is correct because solving the equation with absolute values results in two valid solutions for x: -6 and $(\sqrt{73}-1)/2$.

Option D (None of these) is correct because the solution to the equation is x=1, which is not listed in the options.

Option B $((-\infty,-4))$ is correct because solving the inequality $(x-1)^2(x+4) < 0$ results in this interval.

Option A (0) is correct because the given condition implies m > 3, and the expression inside the greatest integer function is less than 1, resulting in 0.

Option A ((0,1)) is correct because by constructing a function f(x) and applying Rolle's theorem, we can show that at least one root of the given equation lies in the interval (0,1).

The correct answer is A (1) because the minimum value of the expression is found by completing the square after using the discriminant condition for equal roots.

Option A $\left( { - \infty ,\; - 2} \right) \cup \left( { - \frac{1}{2},\;\infty } \right)$ is correct because it represents the solution set of the inequality considering both cases when $x < 1$ and $x \ge 1$.

Option D ($4+\sqrt{5}$) is correct because the harmonic mean formula is applied correctly, and the values are substituted to find the value of b.

Option D (None of these) is correct because the equation has no real roots as the expression ${t^2}{x^2} + \left| x \right| + 9$ is always positive for all real values of x.

Option B $\left( { - \frac{{1}}{{{{13}}}},\\;\frac{{1}}{{{3}}}} \right)$ is correct because the range of the expression is found by ensuring the discriminant of the quadratic equation is non-negative.

Option D (Never less than ${n^2}$) is correct because, by the AM-GM inequality, the sum of the numbers is always greater than or equal to ${n^2}$.

Option A ($x_1=x_3$ and $x_2=x_4$) is correct because by applying AM-GM inequality and the given equations, it can be shown that $x_1 = x_3 = 2$ and $x_2 = x_4 = 1/2$.

Option D ($a=6, b=-4$) is correct because by assuming all roots are equal and positive, we can find the values of a and b that satisfy the given conditions.

Option D ($2i$) is correct because it is the other root of the quadratic equation, found by using the sum of roots property.

Option B ({1,9,1/81}) is correct because these values satisfy the given equation after taking logarithms and solving for x.

Option C ([-2,8]) is correct because it represents the interval of p values for which the discriminant of the given quadratic equation is non-negative, ensuring real roots.

Option B ($0,4$) is correct because solving the equation with absolute values leads to these two solutions.

Option D (None of these) is correct because the given equation has no roots, as any potential solution leads to a contradiction.

Option C ($-1 < m < 3$) is correct because it satisfies the condition that both roots of the quadratic equation are greater than -2 and less than 4.

The correct answer is A ($x^3-3 x^2+15 x-69=0$) because substituting $\alpha = \frac{z-1}{2}$ into the original equation and simplifying yields the new equation.

Option B ($8\sqrt{5}$) is correct because it represents the absolute difference of the cubes of the roots when m=0, which maximizes the sum of the roots.