Quadratic equations practice · Easy

Option A (AP) is correct because the given condition leads to the relationship $2a^2c = ab^2 + bc^2$, which is the condition for $bc^2, ca^2, ab^2$ to be in Arithmetic Progression.

Option A (${p^3} - \left( {3p - 1} \right)q + {q^2} = 0$) is correct because it represents the relationship between p and q when one root is the square of the other root in the given quadratic equation.

Option A (0) is correct because the expression $a A_{n+2}+b A_{n+1}+c A_n$ simplifies to 0 using the properties of roots of a quadratic equation.

Option D (25) is correct because by solving the equations derived from the repeated root condition and the given quadratic equations, we find the value of ${\alpha ^2} + {\beta ^2}$ to be 25.

Option A (0) is correct because the equation has no real roots since the only possible solution leads to ${x^2} = - 1$, which has no real solutions.

Option C ($\left( { - \infty ,\; - 3} \right)$) is correct because it represents the intersection of the solution sets of the two inequalities, $x < 9$ and $x < -3$.

Option B (${x^2} < 2$) is correct because substituting y = a/b into the equation for x and using the given condition ${y^2} > 2$ leads to ${x^2} < 2$.

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

Option D ($\frac{9}{4}\left(9-p^{2}\right)$) is correct because it is derived by using the relationships between the roots and coefficients of the given quadratic equations and simplifying the expression.

Option B (8) is correct because it is the only solution that satisfies the equation $2x - 1 = |x + 7|$.

Option A (0) is correct because if the roots are equal in magnitude but opposite in sign, their sum is zero, which relates directly to the coefficients of the quadratic equation.

Option A ($\left( { - 1,\frac{{ - 1}}{3}} \right)$) is correct because it represents the interval obtained by solving the inequality $\left| {3x + 2} \right| < 1$.

Option C $-1$ is correct because it is the value of $\lambda$ calculated using the vertex form of the quadratic equation and the given highest point.

Option A (0) is correct because the equation has no solution as the LHS is always less than or equal to 2, while the RHS is always greater than or equal to 2, and they equal 2 at different x values.

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

Option C ($\frac{{{b^2} + 4ac}}{{{a^2}}}$) is correct because the difference between the roots remains constant even when the roots change by the same quantity.

Option A (4) is correct because solving for k and checking the discriminant condition gives k=4.

Option C ($a, b$) is correct because $a$ and $b$ are the roots of the equation $(x-\alpha)(x-\beta)+c=0$ as derived from the given equation $(x-a)(x-b)=c$.

Option C ($\pm \;6$) is correct because it is the value of p that satisfies the condition where the difference between the roots of the given quadratic equation is 2.

The speeds of the car and bus are 50 km/hr and 25 km/hr respectively, which is obtained by setting up an equation based on the given time difference and solving for the speeds. Therefore, the correct answer is B ($25 {~km} / {hr}, 50 {~km} / {hr}$).

Option A (Imaginary) is correct because the discriminant of the quadratic equation is negative, implying imaginary roots.

Option C (20) is correct because after finding the equation of the plane and comparing it with the given form, the value of (a-b+c) is calculated to be 22.

Option B ($\alpha = 1\,\mathrm{and}\,\beta = - 2$) is correct because by substituting the roots into the equation and solving the resulting system of equations, we find these values for alpha and beta.

Option B (a+b=1, 0 < b < 1) is correct because using the tangent addition formula and the given conditions, we can derive the relationship a + b = 1 and the constraint on b.

Option D (None of these) is correct because the roots provided are not correct.

Option A (1) is correct because the equation has only one real solution, x = 7/3, after excluding x = ±2.

Option A (4) is correct because substituting x = -1 into the polynomial and setting it to zero allows us to solve for p.

Option D is correct because the set of possible values of 'a' for which both the sum and product of the roots are less than 1 is the interval (1, 5/2).

Option A ($ac{x^2} - bx + 1 = 0$) is correct because it is the transformed equation obtained by replacing $x$ with $\frac{{1 - bx}}{{ax}}$ in the original equation.

Option A (0) is correct because the range of the left side is less than or equal to 2, while the range of the right side is greater than or equal to 2, and they never equal each other except at a single point that doesn't satisfy the equation.