Roots of a Quadratic Equation

Correct answer: A. −2/3

Explanation: Vieta’s product relation states that if α and β are the roots of ax² + bx + c = 0, then αβ = c/a. In the given equation, the leading coefficient is a = 6 and the constant term is c = −4. Substituting these values gives αβ = −4/6 = −2/3. Thus option A is correct. The negative sign must be retained; ignoring it would lead to option B. The values −5/6 and 5/6 are based on the middle coefficient and confuse the product relation with the sum relation. In fact, α + β = −b/a = −5/6, whereas αβ = c/a = −2/3. Therefore the constant and leading coefficients, rather than the middle coefficient alone, determine the required product.

Correct answer: A. \(k\le9\)

Explanation: A quadratic equation has real roots when its discriminant satisfies \(D=b^2-4ac\ge0\). Here, \(a=1\), \(b=6\), and \(c=k\), so \(D=36-4k\). Therefore, \(36-4k\ge0\) gives \(k\le9\). The value \(k=9\) is only the special case of equal roots, whereas \(k>9\) produces non-real roots. Exam tip: For real roots, begin by imposing the condition \(D\ge0\).

Correct answer: A. Both negative

Explanation: The roots are (-6) and (-7) because ((x+6)(x+7)=0). When the sum is negative and product is positive, both roots are negative.

Correct answer: A. -10

Explanation: For the quadratic equation \(x^2+ax+25=0\), the sum of the roots is \(-a\). The given sum is \(5+5=10\), so \(-a=10\), which gives \(a=-10\). Therefore, option A is correct. Exam tip: In \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\).

Correct answer: A. \(\frac{3}{2}\) and \(\frac{1}{5}\)

Explanation: Factor the quadratic: \(10x^2-17x+3=(2x-3)(5x-1)\). Thus, \((2x-3)(5x-1)=0\) gives \(x=\frac{3}{2}\) or \(x=\frac{1}{5}\), so option A is correct. Option B has the signs of both roots wrong. Exam tip: Set each linear factor equal to zero, then verify the roots using the sum and product of roots if needed.

Correct answer: A. \(-c^2\)

Explanation: For the quadratic equation \(ax^2+bx+d=0\), the product of the roots is \(\frac{d}{a}\). Here, the product is \(c\times(-c)=-c^2\), so \(\frac{d}{a}=-c^2\). Option B misses the negative sign. Exam tip: For \(ax^2+bx+d=0\), remember that the sum of roots is \(-\frac{b}{a}\) and the product is \(\frac{d}{a}\).

Correct answer: A. \(x^2+7x-18=0\)

Explanation: If \(\alpha\) and \(\beta\) are the roots of a monic quadratic equation, its form is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). Substituting the given values gives \(x^2-(-7)x+(-18)=0\), which simplifies to \(x^2+7x-18=0\). Option B incorrectly uses the sign of the sum. Exam tip: the coefficient of \(x\) is the negative of the sum of the roots, while the constant term is their product.

Correct answer: A. 53

Explanation: Let the roots be \(\alpha\) and \(\beta\). By the relations between the roots and coefficients, \(\alpha+\beta=9\) and \(\alpha\beta=14\). Hence, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=9^2-2(14)=81-28=53\). Therefore, the correct answer is 53. Exam tip: use \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\); \(81\) alone is only \((\alpha+\beta)^2\), not the required sum.

Correct answer: A. 2

Explanation: The governing concept is the root condition: if x = 4 is a root of an equation, substituting x = 4 must make the left-hand side equal to zero. Substitute into ax² − 10x + 8 = 0: a(4)² − 10(4) + 8 = 0. This becomes 16a − 40 + 8 = 0, or 16a − 32 = 0. Hence 16a = 32 and a = 2. Therefore option A is correct. Checking the alternatives confirms this: a = 1 gives 16 − 40 + 8 = −16; a = −2 gives −32 − 40 + 8 = −64; and a = −1 gives −16 − 40 + 8 = −48. None of those makes the expression zero, so direct substitution establishes a unique answer.

Correct answer: A. 22 and 121

Explanation: For the standard form \(ax^2+bx+c=0\), we have \(a=1\), \(b=-22\), and \(c=121\). The sum of the roots is \(-b/a=22\), and their product is \(c/a=121\). In fact, \(x^2-22x+121=(x-11)^2\), so both roots are 11. Option B has the wrong sign for the sum. Exam tip: use \(-b/a\) for the sum and \(c/a\) for the product.

Correct answer: A. (\frac{17}{4})

Explanation: Here (\alpha+\beta=\frac{5}{4}) and (\alpha\beta=-\frac{3}{2}). Therefore (\alpha+\beta-2\alpha\beta=\frac{5}{4}+3=\frac{17}{4}).

Correct answer: A. Sum is (a+2) and product is (2a)

Explanation: For a monic quadratic x² + bx + c = 0 with roots r and s, Vieta’s relations are r + s = −b and rs = c. In the given equation, the coefficient of x is −(a + 2), so the sum of the roots is −[−(a + 2)] = a + 2. The constant term is 2a, so the product is 2a. Using the stated roots directly gives 2 + a = a + 2 and 2 × a = 2a, confirming option A. The roots are not necessarily equal; they are equal only for a particular value of a. The discriminant is not always negative, and the equation can be factored as (x − 2)(x − a) = 0. Thus option A correctly identifies both the sum and product relations.

Correct answer: A. 10

Explanation: Let the equal roots be r and r. By Vieta’s product relation for the monic equation x² + px + 25 = 0, r² = 25, so r = 5 or r = −5. The condition that both roots are negative selects r = −5. Their sum is therefore r + r = −10. Vieta’s sum relation says that the sum of the roots is −p, so −p = −10 and p = 10. Thus option A is correct. This result is also verified by writing the equation as (x + 5)² = x² + 10x + 25, which shows p = 10 and the repeated root −5. Option B would give repeated positive roots 5 and 5. Options C and D do not satisfy both equality and the required product 25.

Correct answer: A. 10

Explanation: Factor the quadratic first: x² + 4x − 21 = (x + 7)(x − 3), because the two numbers −7 and 3 have sum −4 and product −21. Therefore the roots are −7 and 3. Their absolute difference is |α − β| = |3 − (−7)| = |10| = 10, regardless of which root is named α or β. The discriminant provides an alternative check: Δ = b² − 4ac = 4² − 4(1)(−21) = 16 + 84 = 100. For a monic quadratic, (α − β)² = Δ, so |α − β| = √100 = 10. Hence option A is correct. The other numbers are coefficients or individual root magnitudes, not the distance between the roots.

Correct answer: C. 13

Explanation: Let the positive integer roots be r and s, with r>s. Vieta’s product relation gives rs=36, and the stated difference gives r-s=5. The positive factor pairs of 36 are (1,36), (2,18), (3,12), and (4,9). Only 9-4=5, so the roots must be 9 and 4. Their sum is 9+4=13. For x^2-px+36=0, Vieta’s sum relation says r+s=p, because the coefficient of x is -p. Therefore p=13, making option C correct. The other options do not equal the sum of the only positive factor pair whose difference is 5. This also confirms that the quadratic is x^2-13x+36=(x-9)(x-4).

Correct answer: A. 3 हमेशा एक जड़ है

Explanation: The governing concept is the factor theorem and verification of a proposed root. Let f(x)=x^2−(a+3)x+3a. Substituting x=3 gives f(3)=9−3(a+3)+3a=9−3a−9+3a=0, and the cancellation holds for every value of a. Hence 3 is always a root. In fact, the polynomial can be factorised as x^2−(a+3)x+3a=(x−3)(x−a), so the other root is a. This factorisation also disproves the other choices: a can itself be a root, the roots are equal only when a=3, and for every real a both roots 3 and a are real. Therefore option A is the only statement that remains true without imposing an extra condition on a.

Correct answer: A. 3 और -6

Explanation: The governing concept is factorisation of a quadratic. For x^2+3x−18, we need two numbers whose product is −18 and whose sum is 3, because the constant term is −18 and the coefficient of x is 3. The required numbers are 6 and −3. Thus x^2+3x−18=(x+6)(x−3). Setting each factor equal to zero gives x=−6 and x=3, so the root pair is 3 and −6 and option A is correct. Option B reverses the signs and would correspond to factors (x−3)(x+6) only if the listed values were 3 and −6, not −3 and 6. The pairs in C and D have products −18, but their sums are −7 and 7 respectively, not 3; hence they cannot satisfy the equation.

Correct answer: A. (p + 2)/2

Explanation: The governing concept is the relation between the roots and coefficients of a quadratic equation. For ax² + bx + c = 0, the product of the roots is c/a. Here a = 2 and c = p(p + 2), so the product of the two roots is p(p + 2)/2. If the known root is p and the other root is r, then pr = p(p + 2)/2. For p ≠ 0, cancellation gives r = (p + 2)/2. The same result follows by factoring: 2x² − (3p + 2)x + p(p + 2) = (x − p)(2x − p − 2). Thus the roots are p and (p + 2)/2, making option A correct. The other choices do not satisfy the product-of-roots relation.

Correct answer: A. 8 is always a root

Explanation: To test whether a constant value is always a root, substitute it into the equation. Put x = 8 in x² − (a + 8)x + 8a. The result is 8² − (a + 8)·8 + 8a = 64 − 8a − 64 + 8a = 0, for every value of a. Therefore, x = 8 is always a root. In fact, the expression factors as x² − (a + 8)x + 8a = (x − 8)(x − a), so the two roots are 8 and a, with possible repetition when a = 8. Hence option A is correct. Option B confuses the sum of the roots with a root, option C is true only when a = 8, and option D is false because the roots are real for real a.

Correct answer: A. λ > 9/7

Explanation: For a quadratic equation ax² + bx + c = 0 to have non-real roots, its discriminant must be negative: Δ = b² − 4ac < 0. In this equation, a = 7, b = −6, and c = λ. Therefore, Δ = (−6)² − 4(7)(λ) = 36 − 28λ. The required condition is 36 − 28λ < 0. Subtracting 36 gives −28λ < −36, and dividing by the negative number −28 reverses the inequality, yielding λ > 36/28 = 9/7. Thus option A is correct. Equality λ = 9/7 gives Δ = 0 and equal real roots, while λ < 9/7 gives Δ > 0 and two distinct real roots; hence options B, C, and D are incorrect.

Correct answer: A. x = −1/7, −1

Explanation: The governing concept is factorisation followed by the zero-product property. We need two factors whose product is 7x² + 8x + 1. Splitting the middle term gives 7x² + 7x + x + 1, which factors as (7x + 1)(x + 1). Therefore the equation becomes (7x + 1)(x + 1) = 0. By the zero-product property, either 7x + 1 = 0 or x + 1 = 0. These equations give x = −1/7 and x = −1, respectively. Hence option A is correct. Options B and D lose the negative signs, while option C incorrectly treats 7 as a root instead of solving the linear factor 7x + 1 = 0. Substitution confirms both values make the original expression zero.