Quadratic equations practice · Mixed difficulty

Option A (0,4) is correct because these are the solutions obtained by solving the equation after substituting for the absolute value expression.

Option B ($a{x^2} + 3bx + c = 0$) is incorrect. Option A ($a{x^2} + 3bx + 3c = 0$) is correct because substituting $x/3$ into the original equation and simplifying gives the new quadratic equation with roots three times the original.

Option B (Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1) is correct because both statements are individually true, but Statement-2 does not explain Statement-1.

Option C ($3 / 2$) is correct because it is the value of x that satisfies the given exponential equation after simplification and comparison of exponents.

Option D ($\frac{{ - 2bc}}{{b + c}}$) is correct because solving the equation and using the condition that the product of roots is zero leads to the given sum of roots.

Option B (GP) is correct because the given condition implies that the ratio of the roots is the same, leading to a relationship between the coefficients that satisfies the condition for a geometric progression.

Option D ($p = 1, q = - 1$) is correct because it is derived from finding the sum and product of the new roots $a-2$ and $b-2$, where $a$ and $b$ are roots of ${x^2} - 3x + 1 = 0$.

Option B (1) is correct because the equation simplifies to |x-1|=0, resulting in only one real solution.

Option B (1) is correct because after simplifying the equation and solving for x, only one real root remains after excluding x=1.

Option C (12,8) is correct because the equation can be factored and solved for integer values of x and a.

Option C ($A = G$) is correct because the arithmetic mean of the roots of the first equation equals the geometric mean of the roots of the second equation.

Option C (greater than 2) is correct because the magnitude of the complex roots is greater than 1, and the sum of their magnitudes is greater than 2.

Option C $((-\infty, 3))$ is correct because the condition for at least one negative root requires $a < 3$.

The correct answer is 4 because solving the inequality ${x^2} - 3x - 4 < 0$ gives the range -1 < x < 4, which contains 4 integers.

Option A (4,4) is the correct solution because it satisfies the given equation when substituted for x and y, derived from completing the square and discriminant analysis.

Option C ($\frac{1}{4}$) is correct because it is the minimum value of m that ensures the given expression is always non-negative, derived from the discriminant condition of a quadratic.

Option B (3) is correct because solving the inequality yields -1 < x < 3, which has three integer solutions: 0, 1, and 2.

Option C (No solution) is correct because the expression ${x^2} + 15|x| + 14$ is always positive for all real values of x, so it can never be equal to zero.

Option A (0) is correct because the expression simplifies to zero after substituting the sum and product of the roots.

Option D (All of these) is correct because all the given expressions are equivalent to the other root $\beta$ of the quadratic equation.

Option A ($4 x^2-12 x-7=0$) is correct because given the sum and difference of the roots, we can find the roots and then construct the quadratic equation.

Option C is correct because by using the condition for a common root in quadratic equations, we derive the relationship $a + 4b + 4c = 0$.

Option D ($\frac{p}{q}$) is correct because it is derived by expressing the sum of the reciprocals of the roots in terms of the sum and product of the roots, which are related to the coefficients of the quadratic equation.

Option C (3) is correct because setting up the relationships between the roots and coefficients, and solving for p, yields p = 3.

Option B ($p = 1$ or 0) is correct because these are the possible values of p that satisfy the given conditions for the roots of the equation.

Option B (-13/31) is correct because it is the value obtained by using the conjugate root theorem and Vieta's formulas to find p and q, then substituting them into the given expression.

Option D (40) is correct because it involves finding a common root and using the difference of squares factorization.

Option A ($a = 4,\;b = 4$) is correct because the product $ab$ is maximized when $a$ and $b$ are as close as possible, which occurs when $a = b = 4$.

Option D 2 is correct because it is found by solving the logarithmic equation step by step.

Option D ($- \frac{{a+2b}}{{2a+b}}$) is correct because by simplifying and factoring the given equation, we find that one of the roots is $- \frac{{a+2b}}{{2a+b}}$.