Quadratic equations practice · Easy

Option B (-2) is correct because subtracting the two equations leads to x=1 as a common root, which can then be used to solve for a+b.

Option A (0) is correct because the left-hand side of the equation is always positive, while the right-hand side is always negative, so there are no real solutions.

Option A (7) is correct because the given expression can be simplified into a quadratic equation, and 7 is the positive root of that equation.

Option B ($2\sqrt {ab} < a + b$) is correct because the Arithmetic Mean is always greater than or equal to the Geometric Mean for distinct positive numbers.

Option B ($(x - 4)(x - 9) \ge 0$) is incorrect. Option A ($(x - 4)(x - 9) \le 0$) is correct because for $x$ between 4 and 9 inclusive, $(x-4)$ is non-negative and $(x-9)$ is non-positive, making their product non-positive.

Option C ($p$ is an even integer) is correct because the given condition leads to $p = 2|s|$, which implies that $p$ is an even integer.

Option C ($a+b$ is always an integer) is correct because if $x^2+ax+b$ is an integer for every integer $x$, then evaluating at $x=0$ and $x=1$ shows that $b$ is an integer and $a+b$ is an integer.

Option B (2) is correct because substituting |x| = y results in a quadratic equation with two positive solutions for |x|, leading to two real solutions for x.

Option A ($p=1, q=-2$) is correct because these values satisfy both the sum and product of roots conditions for the given quadratic equation.

Option B ($4\;ac > {b^2}$) is correct because it ensures the quadratic expression always has the same sign as c by guaranteeing no real roots.

Option B (5) is correct because the product of the roots of a quadratic equation is c/a, and if one root is the reciprocal of the other, their product is 1.

Option B ($9 x^2+2 x+1=0$) is correct because it is obtained by finding the sum and product of the new roots ($\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$) and constructing the quadratic equation.

Option A (4) is correct because the equation has four solutions: $x = \pm 2$ and $x = \pm 3$.

Option A (1) is correct because the minimum value of the sum of the distances from z to 0 and z to 1 is the distance between 0 and 1, which is 1.

Option B (47) is correct because using the relationship between roots and coefficients and simplifying the expression yields the value 47.

Option C ($\sqrt{3}$) is correct because it is the absolute value of b, derived from solving the equations with a common root.

Option D ($x \in \left[ { - 2,\; - 1} \right) \cup \left[ {3,\;4} \right)$) is correct because it represents the intervals where the greatest integer function satisfies the given equation.

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

Option B (0) is correct because the condition for common roots simplifies to $2r - p = 0$.

Option A ($a^2-b^2+2 a c=0$) is correct because it is the result of applying Vieta's formulas to relate the roots (sin α, cos α) to the coefficients of the quadratic equation and then using the trigonometric identity sin²α + cos²α = 1.

Option B $\ge 2$ is correct because it is derived using the AM-GM inequality.

Option C (0) is correct because the equation has no solution since the only possible solution, x=1, is an extraneous solution.

Option B (50) is correct because the integral simplifies to 50 after applying integration by parts and simplification.

Option C ($\frac{{q' - q}}{{p - p'}}$) is correct because it represents the common root derived from subtracting the two quadratic equations.

Option A ($7 / 8$) is correct because it is the minimum value of the quadratic equation calculated using the formula $\frac{{4AC-B^2}}{{4A}}$.

Option A ($b+6c=0$) is correct because, since a, b, and c are real, the complex conjugate of the root must also be a root, allowing us to find relationships between the coefficients.

Option A ($x - axis$) is correct because simplifying the given equation leads to y = 0, which represents the x-axis.

Option A (Real and unequal) is correct because the discriminant of the quadratic equation is positive, indicating two distinct real roots.

Option D ($a/b = c/d$) is correct because the condition for equal roots leads to the equation $(ad-bc)^2 = 0$, which simplifies to $a/b = c/d$.

The correct answer is A ($- 5 < a < 2$) because for the quadratic to be always positive, the discriminant must be negative.