Quadratic equations practice · Mixed difficulty

Option C $\left( {1000,\infty } \right)$ is correct because it is the solution set obtained after solving the logarithmic inequality.

Option A (${a^2},{c^2},{b^2}$ are in A.P.) is correct because it is derived from the condition that the given quadratic equations have a common root, and $a, b, c$ are in A.P.

Option A (Minimum value 81) is correct because the minimum value of ${x^3} + {y^3} + {z^3}$ is found using the power mean inequality given ${x^2} + {y^2} + {z^2} = 27$.

The correct answer is D ($5{S_6} + 6{S_5} = 2{S_4}$) because it directly applies Newton's Theorem for recurrence relations of roots of a polynomial equation.

Option B (${q^2} - {p^2}$) is correct because by substituting the roots into the given expression and using the relationships between the roots and coefficients of the quadratic equations, the expression simplifies to the difference of squares of q and p.

Option B ($\left(0, \frac{{\pi}}{{6}}\right) \cup\left(\frac{{5 \pi}}{{6}}, \pi\right)$) is correct because the roots having opposite signs implies that $f(0) < 0$, which leads to the given interval for $\theta$.

The correct answer is B (11) because it is the value of n for which the quadratic equation has solutions α and α+1.

Option C ($x \geq 3$) is correct because the absolute value equation |3-x| = x-3 is only true when x-3 is greater than or equal to zero, which means x must be greater than or equal to 3.

Option B ($- \frac{1}{2}\left( {{{{a^2} + {b^2}}}} \right)$) is correct because it represents the product of the roots under the given condition that the sum of the roots is zero.

Option B (real and both negative) is correct because the roots of the resulting quadratic equation are -7/3 and -1/2, both real and negative.

Option B ($n\left( {n + 1} \right),n \in Z$) is correct because the discriminant must be a perfect square, which leads to m being of the form n(n+1).

Option A is correct because by using Vieta's formulas and simplifying the expression, the value is equal to 2.

Option A is correct because by finding the derivative of f(x) and substituting it into the given equation, we can solve for x, which yields x = $\pm$1.

Option B (16) is correct because by finding the value of lambda using the even prime root and substituting it into the second equation, we can find the value of mu using the discriminant condition for equal roots.

Option A ($a^2+b^2$) is correct because the maximum value of the given expression is obtained by completing the square and recognizing that the squared term is always non-negative.

Option B (2) is correct because the function is not differentiable at x=1 and x=4 due to the presence of absolute value functions creating sharp edges.

Option A (1) is correct because the difference between the roots is 1, and substituting the sum and product of roots into $b^2 - 4c$ simplifies to 1.

Option A ($p = 2, q = 32$) is correct because the values satisfy the conditions of the increasing geometric progression formed by the roots of the given quadratic equations.

Option B is correct because the discriminant of the quadratic equation is greater than 0, implying real roots.

Option C (0) is correct because the equation simplifies to a form where no real value of x satisfies the condition.

Option D (-2/a) is correct because simplifying the given expression using Vieta's formulas leads to -2/a.

Option A (01:02:03) is correct because if two quadratic equations have a common root and one has imaginary roots, then both roots must be common, implying proportional coefficients.

Option C ($\frac{{ - 9}}{8}$) is correct because it represents the minimum value of the quadratic function, found by completing the square or using the vertex formula.

The correct answer is A ($p=1, q=-2$) because substituting these values into the equations $p+q=-p$ and $pq=q$ derived from Vieta's formulas satisfies the conditions.

Option A ($p$ and $q$ respectively) is correct because the inequalities lead to the conclusion that p is the largest and q is the smallest.

Option B (16) is correct because it is the value of $\mu$ obtained by using the given information that one root of the first equation is 2 and the second equation has equal roots.

Option C ($a^4 x^2-2 a^2(b^2-2 c a) x+b^2(b^2-4 c a)=0$) is correct because it is derived by finding the sum and product of the squares of the sum and difference of the roots of the given quadratic equation.

Option A (1) is correct because by setting up the equations based on the given information about the roots and simplifying, we find that ${b^2} - 4c$ equals 1.

Option C ([{-2, 0}] $\cup$ [{2, 4}]) is correct because the intersection of the solution sets for the two inequalities |x - 1| <= 3 and |x - 1| >= 1 is the union of the intervals [{-2, 0}] and [{2, 4}].

Option A (1) is correct because the curves $y = {e^{x - 8}}$ and $y = 17 - 2x$ intersect at only one point, (8, 1), indicating one real root.