Quadratic equations practice · Mixed difficulty

Option A ($\{\frac{{\pi}}{{4}}\}$) is correct because it is the only solution to the given logarithmic equation within the specified interval.

Option B ($x^2+4 a x+2 b=0$) is correct because it is derived by substituting the sum and product of the roots into the given expression and simplifying.

Option A ($\left[-1, \frac{{-1+\sqrt{5}}}{{2}}\right)$) is correct because by considering the domain and solving the inequality, we find the solution set to be $\left[-1, \frac{{-1+\sqrt{5}}}{{2}}\right)$.

Option C (0) is correct because substituting the roots α and β into the given expression results in zero, due to the properties of the roots satisfying the original quadratic equation.

Option C (${\left( {{a^3}c} \right)^{1/4}} + {\left( {a{c^3}} \right)^{1/4}} + b = 0$) is correct because it is derived from the given condition $\beta = {\alpha ^{1/3}}$ and the relationships between the roots and coefficients of the quadratic equation.

Option A ($a=b=c$) is correct because if all three quadratic equations have only positive roots, it implies that the coefficients must be equal.

Option B ($p=1$ or 0) is correct because substituting the roots into the given equation and solving the resulting system of equations yields these two possible values for p.

Option A (254) is correct because substituting the given values and simplifying the expression results in 254.

Option C ($\frac{{29}}{{348}}$) is correct because by finding the sum of the infinite geometric series formed by the roots and using the relationships between the roots and coefficients, the limit simplifies to the given fraction.

Option A (0) is correct because the equation has no solution as the value of sin x obtained is greater than 1.

Option C (1, -4) is correct because by finding the values of 'b' and 'c' from the given roots and substituting them into the equation x² + bx + c = 0, we can solve for the roots.

The correct answer is C. By analyzing the sign changes in the polynomial and evaluating the function at specific points, we can conclude that there is only one positive real root.

Option D ($\{125, 8\}$) is correct because both 125 and 8 satisfy the given equation.

Option A (A.P.) is correct because if the given equations have a common root, then $\frac{d}{a}$, $\frac{e}{b}$, $\frac{f}{c}$ must be in Arithmetic Progression.

Option D ($\frac{{ - 71}}{{256}}$) is correct because by using the centroid coordinates and the arithmetic progression property, we can find the values of a, b, and c, and then calculate the required expression using the roots of the quadratic equation.

Option C ($-2 < x < -1$) is correct because solving the inequality leads to the solution set where x is between -2 and -1.

Option B (5/2) is correct because after solving for the roots of the equation, the product of the real roots is calculated to be 5/2.

Option A (2) is correct because the least value of |a| for which tan θ and cot θ are roots of the equation is 2, derived from the condition for real roots.

Option A ($- 2, - 32$) is correct because these values satisfy the given conditions for the roots to be in GP.

Option C ($\frac{1}{2}$) is correct because this value of $\lambda$ minimizes the sum of the squares of the roots of the given quadratic equation.

Option C (${p^2} = q(q + 2)$) is correct because by using the sum and product of roots, and the trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$, we can derive the given relation.

Option C (5) is correct because using the given conditions $2b = a+c$ and $a^2 + b^2 + c^2 = 84$, and the triangle inequality, we can deduce that b = 5.

Option C ($1+\log _8(6)$) is correct because by substituting $8^x = t$, solving the quadratic equation, and summing the logarithmic solutions, we arrive at the given answer.

Option B ($1 + \sqrt 2$) is correct because it is the positive solution to the quadratic equation formed by the infinite continued fraction.

Option A ($k = \pm 4$) is correct because these values satisfy the condition for the two quadratic equations to have a common root, found by solving the system of equations.

Option A (at least one root in [0,1]) is correct because Rolle's theorem guarantees the existence of a root within the interval (0,1) for the derivative of a function that is zero at both endpoints.

Option A $\left[ {2/3,\;2} \right]$ is correct because it represents the range of possible values for $x$ given the constraints on $x, y,$ and $z$.

Option A (0, 24) is correct because these are the values of 'a' that result in a common root for the two quadratic expressions.

Option D (None of these) is correct because the value of the expression is 4, which is not among the options.

Option A ($1 \le x \le 3$) is correct because by completing the square and analyzing the resulting equation, we can determine the range of possible values for x.