Quadratic equations practice · Easy

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

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 $- \frac{{a+2b}}{{2a+b}}$.

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

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

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 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.

The correct answer is D ($\left\{ {\frac{{p + q}}{p},\;\frac{{p + q}}{q}} \right\}$) because these are the solutions obtained by solving the given quadratic equation using the quadratic formula or factoring.

The correct answer is B ($2 \alpha, 2 \beta$) because the new roots are twice the original roots, which can be derived by manipulating the quadratic equation.

Option D ($\frac{2}{9}(2p - q)(2q - p)$) is correct because solving the system of equations derived from Vieta's formulas yields this expression for r.

Option D (9) is correct because the least value of $a + b + c$ is 9 when the two quadratic equations have a common root and $a, b, c$ are natural numbers.

The correct answer is D. The equation has no solution because the LHS is always less than or equal to 1, while the RHS is always greater than or equal to 2.

Option C (Real) is correct because the discriminant of the quadratic equation is always non-negative, ensuring that the roots are real.

Option C (&nbsp;-2) is correct because the condition that one root exceeds 2 and the other is less than 2 implies $f(2) < 0$, which leads to the solution $a < -2$.