Roots of a Quadratic Equation

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.

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 \(D\ge0\).

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.

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 \(c\).

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.

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 \(\frac{d}{a}\).

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.

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.

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.

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.

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}).

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.

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.

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.

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).

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.

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.

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.

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.

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.

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.