Quadratic Equations MCQ Test

For 2x² − kx + 1 = 0, equal roots require the discriminant to be zero. Here, D = (−k)² − 4(2)(1) = k² − 8. Setting this equal to zero gives k² = 8. Taking square roots yields k = ±2√2. Both values ensure the discriminant is zero, producing equal real roots for the given quadratic equation.

Factor the first equation: x² − 3x + 2 = (x − 1)(x − 2), giving roots 1 and 2. Factor the second equation: 2x² − 5x + 2 = (2x − 1)(x − 2), giving roots 1/2 and 2. The common root appearing in both sets is x = 2. Substitution confirms it satisfies both equations simultaneously.

The quadratic x² − px + 12 = 0 has no real roots when its discriminant is negative. The discriminant is D = p² − 48. For no real roots, p² − 48

Given x² − 5x + 1 = 0 and x ≠ 0, divide the equation by x to obtain x − 5 + 1/x = 0. Rearranging gives x + 1/x = 5. This method avoids solving for x explicitly and directly finds the required expression. The result follows from algebraic manipulation rather than root calculation.

To verify whether a value is a solution, substitute it into x² − 5x + 6 = 0. For x = 3, substitution gives 9 − 15 + 6 = 0, which satisfies the equation. Other values do not reduce the expression to zero. Hence, x = 3 is a valid solution. This confirms the root using direct substitution, a standard method for checking solutions.

Equal roots occur when the discriminant equals zero. For x² + kx + 9 = 0, the discriminant is D = b² − 4ac = k² − 36. Setting D = 0 gives k² = 36, which results in k = ±6. Both values lead to identical roots using the quadratic formula, so ±6 is the correct condition for equal roots in this equation.

For x² − 4x + 4 = 0, the discriminant is D = (−4)² − 4(1)(4) = 16 − 16 = 0. A zero discriminant means the equation has two equal real roots. In fact, the equation factors as (x − 2)² = 0, giving a repeated root at x = 2. This confirms equality of roots clearly.

For x² + 4x + 5 = 0, compute the discriminant D = b² − 4ac = 16 − 20 = −4. Since the discriminant is negative, the equation has no real roots. Negative discriminant values indicate complex conjugate solutions, not real ones. Therefore, the equation does not intersect the x-axis at any real point.

For x² − 5x + k = 0 to have equal roots, the discriminant must be zero. Compute D = (−5)² − 4(1)(k) = 25 − 4k. Setting D = 0 gives 4k = 25, so k = 25/4. This value ensures the roots coincide at x = 5/2, confirming equality of roots mathematically.

A quadratic equation must have degree two, meaning the highest power of x is 2. Among the options, 5x − 7 = 0 has degree one, so it is a linear equation. All other options contain x² either directly or after expansion, which confirms they are quadratic equations. Therefore, the linear equation is the correct answer to identify the non-quadratic expression.

If one root is 2 + 3i and coefficients are real, the conjugate 2 − 3i must also be a root. Form the equation using (x − 2 − 3i)(x − 2 + 3i). This expands to (x − 2)² + 9 = 0, which simplifies to x² − 4x + 13 = 0. This equation has real coefficients and complex roots.

Let the roots of x² − kx + 12 = 0 be r and s. Then r + s = k and rs = 12. Given the difference of roots is 1, (r − s)² = 1. But (r − s)² = (r + s)² − 4rs = k² − 48. Setting k² − 48 = 1 gives k² = 49, so the positive value is 7.

Since x = −2 is a root of the equation x² + kx − 6 = 0, substitute x = −2. This gives (−2)² + k(−2) − 6 = 0, which simplifies to 4 − 2k − 6 = 0. Solving −2k − 2 = 0 gives k = −1. Only this value satisfies the condition that x = −2 makes the equation equal to zero.

Expanding kx(x − 2) + 6 = 0 gives kx² − 2kx + 6 = 0. The discriminant is D = (−2k)² − 4(k)(6) = 4k² − 24k. Setting D = 0 yields 4k(k − 6) = 0. Since k ≠ 0 for a quadratic equation, k = 6 is the only valid solution.

Using Vieta’s formulas for px² + 3x + q = 0 with roots −1 and −2, the sum of roots equals −3 = −3/p, giving p = 1. The product of roots equals 2 = q/p, so q = 2. Adding p and q gives 1 + 2 = 3. This result comes directly from standard relationships between roots and coefficients.