Quadratic equations practice · Easy

Option B (3) is correct because it is the smallest integer greater than the solution to the inequality.

Option A is correct because the roots of the given quadratic equation are rational, as they can be expressed as -1 and -b/2a.

Option A (1) is correct because the equation simplifies to a single valid solution for x.

The correct answer is D ($\frac{{b}}{{ac}}$) because by using Vieta's formulas and simplifying the expression, we arrive at this result.

Option A (3) is correct because the minimum value of the sum of absolute differences occurs at the median, which is 3, and the minimum value is |3-3|+|3-2|+|3-5| = 0+1+2 = 3.

Option C (7, 1/7) is correct because the maximum and minimum values are found by setting the discriminant greater than or equal to zero and solving for y.

Option D (4) is correct because it is the only value of $k$ that satisfies both the product of roots condition and the real roots condition.

Option D (18) is correct because the given conditions allow us to express p(x) as k(x+1)², solve for k using p(-2)=2, and then evaluate p(2).

Option C (4) is correct because simplifying the square root and then adding x and 1/x results in the cancellation of the square root term, leaving 4.

Option A $\left( { - 1,\;1} \right] \cup \left[ {5,\;7} \right)$ is correct because it represents the solution set of the compound inequality involving absolute value.

Option D (-2) is correct because the equation not having two distinct real roots implies that the discriminant is non-negative, leading to the minimum value of $3a+b$.

Option D (${\left( {{a_1}{c_2} - {c_1}{a_2}} \right)^2}$) is correct because the condition for a common root leads to this relationship between the coefficients.

Option A ($\left( {2 - \sqrt 3 ,\;2 + \sqrt 3 } \right) \cup \left( { - 2 - \sqrt 3 ,\; - 2 + \sqrt 3 } \right)$) is correct because it represents the solution set obtained by solving the inequality $\left| {x + \frac{1}{x}} \right| < 4$ by considering both positive and negative cases and finding the intersection of the solution sets.

Option C (1) is correct. By using the given equation and substituting, we can simplify the expression to 1.

Option B ((0, 1)) is correct because for roots to have opposite signs, the product of roots must be negative, leading to the inequality $\frac{{p(p - 1)}}{3} < 0$ which is satisfied when $p \in (0, 1)$.

Option B (1) is correct because after solving the logarithmic equation and checking for extraneous solutions, only one valid solution remains.

Option B (2) is correct because simplifying the exponential equation leads to a quadratic equation with a positive discriminant, indicating two real roots.

Option D (1) is correct because simplifying the expression for x and then substituting into the given expression results in a value of 1.

The correct answer is C (-1) because substituting the common root into both equations and solving leads to a + b = -1.

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 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 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.

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 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.