Roots of a Quadratic Equation

The governing concept is formation of a quadratic equation from its roots. If the roots are α and β, a monic quadratic equation with those roots is (x − α)(x − β) = 0. Here α = 1 and β = 4, so form the factors (x − 1) and (x − 4). Multiplying gives x² − 4x − x + 4 = x² − 5x + 4. Therefore the required equation is x² − 5x + 4 = 0, which is option A. Its root sum is 5 and its root product is 4, agreeing with 1+4 and 1×4. The remaining options have different signs or interchange coefficients and do not produce both required roots.

For a quadratic equation ax² + bx + c = 0, the sum of the roots is −b/a. Comparing 2x² + mx + 8 = 0 with the standard form gives a = 2 and b = m. Therefore the root sum is −m/2. The question states that this sum equals −3, so −m/2 = −3. Multiplying both sides by 2 gives −m = −6, and multiplying by −1 gives m = 6. Thus option A is correct. A common error is to use m/2 without the required negative sign, leading to option B. The constant 8 does not affect the root sum; it determines the product c/a = 4. This confirms that only the coefficient m must be solved here.

The governing idea is factorisation of a quadratic. We need two numbers whose product is −15 and whose sum is −2, because the equation is x² − 2x − 15 = 0. The numbers 5 and −3 satisfy both conditions: 5 × (−3) = −15 and 5 + (−3) = 2; after factorisation, x² − 2x − 15 = (x − 5)(x + 3). Setting each factor equal to zero gives x = 5 or x = −3. Therefore option B is correct. Option A has the same product but sum −2? Actually 3 and −5 sum to −2, but their product is −15; it corresponds to x² + 2x − 15, not the given equation. Options C and D have incorrect sums or products.

The governing concept is the formation of a monic quadratic from its roots. If α and β are the roots, the equation is (x − α)(x − β) = 0. Here α = 2 and β = 7, so we write (x − 2)(x − 7) = 0. On expanding, x² − 7x − 2x + 14 = 0, which becomes x² − 9x + 14 = 0. Therefore option A is correct. The result can also be checked using Vieta’s relations: the sum of roots must be 2 + 7 = 9, so the coefficient of x in a monic equation is −9, while the product must be 2 × 7 = 14, giving the constant term 14. Option B has the wrong sign for the sum, option C interchanges sum and product, and option D has unsuitable signs and coefficients.

The governing principle is the definition of a root: substituting a root into the polynomial must make its value zero. Since x = −3 is given as a root, substitute it into 2x² + rx − 9 = 0. We obtain 2(−3)² + r(−3) − 9 = 0, so 18 − 3r − 9 = 0. Combining constants gives 9 − 3r = 0, hence 3r = 9 and r = 3. Therefore option A is correct. A common error is to treat r(−3) as +3r, but multiplication by a negative number gives −3r. Verification is immediate: with r = 3, the expression is 2(9) + 3(−3) − 9 = 18 − 9 − 9 = 0. Thus the stated value really is a root; the other options fail this substitution test.

To form a monic quadratic when the roots are 3 and 8, use the factor form (x − 3)(x − 8) = 0. Expanding gives x² − 8x − 3x + 24 = 0, and combining like terms produces x² − 11x + 24 = 0. Thus option A is correct. Vieta’s relations provide an independent check: for x² + bx + c = 0, the sum of the roots is −b and their product is c. The sum here is 3 + 8 = 11, so b must be −11; the product is 3 × 8 = 24, so c must be 24. Option B has the wrong sign for the x coefficient, option C reverses the sum and product, and option D has both an incorrect coefficient pattern and an incorrect constant sign.

The sum of two given roots is found by ordinary addition. Here the roots are −6 and 1, so their sum is (−6) + 1 = −5. Therefore option A is correct. The negative sign must be retained because the magnitude of −6 is greater than the magnitude of 1. Option B results from ignoring the negative sign on −6. Option C incorrectly subtracts 1 from −6 instead of adding the second root, while option D uses the wrong sign and magnitude. In terms of a quadratic equation, this result is also consistent with Vieta’s relation: if the roots are α and β, then α + β is the sum of the roots, equal to −b/a for ax² + bx + c = 0.

The governing concept is Vieta’s product relation. For a quadratic equation ax² + bx + c = 0 with roots α and β, the product of the roots is αβ = c/a. Comparing 6x² + 5x − 14 = 0 with the standard form gives a = 6 and c = −14. Therefore αβ = −14/6 = −7/3 after dividing numerator and denominator by 2. Hence option A is correct. There is no need to calculate either root separately. Option B has the correct magnitude but an incorrect positive sign; the negative constant term makes the product negative because a is positive. Options C and D incorrectly use the coefficient b = 5, whereas b determines the sum of the roots, not their product. The relation is valid even when the roots are not explicitly found.

Let the second root be s. For a monic quadratic x² + ax + 18 = 0, the product of the roots equals the constant term, so 2s = 18 and therefore s = 9. The sum of the roots is the negative of the coefficient of x: 2 + 9 = −a. Thus 11 = −a, giving a = −11. The result can also be verified directly by substituting the known root x = 2: 2² + (−11)(2) + 18 = 4 − 22 + 18 = 0. Option B has the wrong second-root sign and coefficient, while option C has the correct second root but the wrong sign of a. Option D does not satisfy the product relation. Hence option A is the only correct pair.

For a monic quadratic with roots α and β, the equation has the form x² − (α + β)x + αβ = 0. Consequently, the constant term is the product of the roots, not their sum. Here the roots are r and 2r, and their sum is 9, so r + 2r = 3r = 9. Hence r = 3, and the roots are 3 and 6. Their product is 3 × 6 = 18, which is therefore the constant term. Option B is the given sum and not the required product. Options C and D do not equal the product of the two roots. The monic condition is important because it ensures that the constant term itself, rather than a scaled version of it, equals the product of the roots. Thus option A is correct.

Let the unknown root be s. Since the coefficient of x² is 1, the equation is monic and the product of its roots equals the constant term. Thus 4s = 24, so s = 6. The sum of the roots equals the negative coefficient of x, giving 4 + 6 = −a. Therefore 10 = −a and a = −10. Substitution verifies the result: for x = 4, we obtain 4² + (−10)(4) + 24 = 16 − 40 + 24 = 0. Option C has the correct second root but gives the wrong sign for a. Options B and D fail the product relation, since their proposed roots do not multiply with 4 to give 24. Hence option A gives both required values and is correct.

A monic quadratic whose roots are α and β is x² − (α + β)x + αβ = 0, so its constant term is the product of the roots. The given roots are r and 3r, and their sum is 16. Therefore r + 3r = 4r = 16, which gives r = 4. The roots are consequently 4 and 12. Their product is 4 × 12 = 48, so the constant term is 48. Option B repeats the sum of the roots rather than calculating their product. Options C and D do not equal 4 × 12 and arise from incorrect arithmetic or from using an irrelevant expression. The word monic is essential because the leading coefficient is one, making the constant term exactly equal to the product of the roots. Hence option A is correct.

For a quadratic x² + bx + c = 0, Vieta’s relations state that the sum of the roots is −b and their product is c. Comparing the given equation x² − 6x − 27 = 0 with this form gives α + β = 6 and αβ = −27. Rewrite the required expression by grouping the linear terms: αβ + 3α + 3β = αβ + 3(α + β). Substitution now gives −27 + 3(6) = −27 + 18 = −9. Therefore option A is correct. Option B is only the product αβ and ignores the remaining terms. Option C has the wrong sign, while option D results from an incorrect use of the root sum and product. No individual roots need to be calculated.

Let the second root be s. Because the leading coefficient is 1, the product of the roots is equal to the constant term 40. Hence 5s = 40, giving s = 8. The sum of the roots is the negative of the coefficient of x, so 5 + 8 = −a. This gives 13 = −a and therefore a = −13. Direct substitution confirms the known root: 5² + (−13)(5) + 40 = 25 − 65 + 40 = 0. Option C has the correct second root but reverses the sign of a. Options B and D do not satisfy the product condition, since their proposed second roots are inconsistent with 5s = 40. Thus option A is the unique correct answer.

For x² + bx + c = 0, Vieta’s relations give α + β = −b and αβ = c. In x² − 7x − 30 = 0, this means α + β = 7 and αβ = −30. The required expression should be reorganized before substitution: αβ + 4α + 4β = αβ + 4(α + β). Using the known values, it becomes −30 + 4(7) = −30 + 28 = −2. Therefore option A is correct. Option B gives only the product and leaves out the two linear terms. Option C is the opposite sign of the correct result, and option D comes from adding the numbers incorrectly rather than multiplying the root sum by 4. Individual roots are unnecessary because the expression is symmetric.

The governing concept is the relationship between the roots and coefficients of a quadratic equation. For A x² + Bx + C = 0 with roots α and β, αβ = C/A and α + β = −B/A. Here the leading coefficient is 1, so the product of the roots is 54. If one root is 6 and the other is β, then 6β = 54, giving β = 9. Their sum is therefore 6 + 9 = 15. In the given equation the coefficient of x is a, so α + β = −a. Hence −a = 15 and a = −15. Thus option A is correct. Option B uses a wrong sign and wrong product, option C reverses the sign of a, and option D does not satisfy the product relation.

The governing concept is Vieta’s relation for a monic quadratic. If x² + bx + c = 0 has roots α and β, then α + β = −b and αβ = c. Comparing the given equation x² − 8x − 33 = 0 with this form gives α + β = 8 and αβ = −33. The required expression should be grouped before substitution: αβ + 5α + 5β = αβ + 5(α + β). Therefore its value is −33 + 5(8) = −33 + 40 = 7. Hence option A is correct. Option B gives only the product αβ, option C introduces an unnecessary sign reversal, and option D results from adding the terms incorrectly. There is no need to calculate α and β separately.

Let the real roots be α and β. If αβ is positive, α and β must have the same sign: either both are positive or both are negative. The additional condition α + β > 0 distinguishes between these two cases. If both roots were negative, their sum would be negative, contradicting the given condition. Therefore both roots must be positive. A positive and a negative root would produce a negative product, so option C is impossible. A zero root would make the product zero, not positive, so option D is also impossible. Hence option A follows solely from the signs of the sum and product.

(p(r)=0) means the equation is satisfied when (x=r). Substitution is the safest way to check a root.

To check whether a number is a root, substitute it into the equation and verify whether the result is zero. For (x=3) , (3^2-7\times3+12=9-21+12=0) , so 3 is a root of the equation. The other root is 4, so option C is incorrect. Exam tip: For root verification, substitute the given value directly and check whether the expression becomes zero.

To test whether a number is a root, substitute it into the equation and check whether the result is zero. For \(x=-1\), \(2(-1)^2+5(-1)+3=2-5+3=0\). Therefore, \(x=-1\) is a root. In exams, carefully check the sign of the linear term and remember that the square of a negative number is positive.

The discriminant \(D=b^2-4ac\) determines the nature of roots. When \(D=0\), \(x=\frac{-b}{2a}\) occurs twice, so the roots are real and equal. Exam tip: check the sign of the discriminant first.

The equation can be factorised as \(2x^2-7x+3=(2x-1)(x-3)\). Therefore, \((2x-1)(x-3)=0\) gives \(x=\frac{1}{2}\) or \(x=3\), so option A is correct. The roots in option D do not have the required sum and product for this equation. Exam tip: verify the roots using sum \(\frac{-b}{a}=\frac{7}{2}\) and product \(\frac{c}{a}=\frac{3}{2}\).

Splitting the middle term gives \(3x^2+10x+3=3x^2+9x+x+3=(3x+1)(x+3)\). Therefore, \((3x+1)(x+3)=0\) gives \(x=-\frac{1}{3}\) or \(x=-3\), so option A is correct. Exam tip: After factorising, set each factor equal to zero and verify that the sum of the roots is \(-\frac{10}{3}\).

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Here, \(a=5\) and \(b=-4\), so the sum is \(-\frac{-4}{5}=\frac{4}{5}\). Exam tip: substitute the signed value of \(b\), not just its magnitude, in the formula.