Roots of a Quadratic Equation

For p(x) = x² + 6x + 7, the coefficients are a = 1, b = 6, and c = 7. The discriminant is D = b² − 4ac = 6² − 4(1)(7) = 36 − 28 = 8. Since D > 0, the quadratic has two distinct real zeroes. Because 8 is not a perfect square, √8 is irrational, so the zeroes are irrational rather than rational. The quadratic formula confirms this: x = (−6 ± √8)/2 = −3 ± √2. Therefore option A is correct. Option B would require a positive perfect-square discriminant, option C would require D = 0, and option D would require D < 0. These conditions do not hold here.

The governing concept is the root condition. A number is a root of a polynomial equation exactly when substituting that number makes the left-hand side zero. Substituting x = 1 into kx² − 5x + 2 gives k(1)² − 5(1) + 2 = k − 3. The statement says that x = 1 is not a root, so this expression must not equal zero. Therefore k − 3 ≠ 0, which gives k ≠ 3. Hence option A is correct. If k were 3, the expression would be 3 − 5 + 2 = 0, making x = 1 a root, contrary to the question. The values k = 5 and k ≠ 5 do not follow from the substitution condition and are irrelevant distractors. The argument does not require solving the quadratic.

Compare the equation x² + 3x − 18 = 0 with the standard form ax² + bx + c = 0. Here a = 1, b = 3, and c = −18. By Vieta’s relations, α + β = −b/a = −3 and αβ = c/a = −18. The required expression can be grouped as αβ − α − β = αβ − (α + β). Substituting the known values gives −18 − (−3) = −18 + 3 = −15. Therefore option A is correct. There is no need to find α and β separately. The other options usually result from forgetting the negative sign in α + β, adding the root sum instead of subtracting it, or incorrectly treating the constant term as positive. The calculation relies only on the coefficients and the root relations.

Here (\alpha+\beta=7) and (\alpha\beta=12). (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{25}{12}).

((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=25-24=1). Remember this identity for difference of roots.

Here, \(a=1\), \(b=-2k\), and \(c=k^2\). The discriminant is \(D=b^2-4ac=(-2k)^2-4(1)(k^2)=0\). Hence the roots are real and equal; equivalently, the equation is \((x-k)^2=0\), so both roots are \(k\). Option D is true only for the special case \(k=0\), not for every value of \(k\). Exam tip: when \(D=0\), a quadratic equation has equal real roots.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, the product of the roots is \(2\times3=6\) and \(c=6\). Thus, \(\frac{6}{a}=6\), giving \(a=1\). Therefore, option A is correct. Exam tip: remember that the sum of roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\).

By Vieta’s theorem, the sum of the roots of x² − (m + 2)x + 12 = 0 is m + 2. Since the given roots are 3 and 4, their sum is 7, so m + 2 = 7 and m = 5. Their product, 3 × 4 = 12, also confirms the constant term. Option B, 7, is the sum of the roots, not the value of m. Exam tip: For x² + bx + c = 0, the sum of the roots is −b.

By Vieta’s formulas, the sum of the roots is \(-p\) and their product is \(q\). Here, the sum is \((-2)+(-5)=-7\), so \(-p=-7\) gives \(p=7\). Their product is \((-2)(-5)=10\), hence \(q=10\). Therefore, \(p+q=7+10=17\). In exams, match the sum and product of the roots with \(-p\) and \(q\), respectively.

For a quadratic equation \(x^2+bx+c=0\), the product of the roots is \(\alpha\beta=c\). Here, \(c=k\), so \(k=(-2)(-4)=8\). The option \(-8\) results from an incorrect sign in multiplying two negative numbers. Exam tip: Verify the roots using both their sum \((-b)\) and product \(c\).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here (\frac{\frac{7}{2}}{\frac{3}{2}}=\frac{7}{3}).

(\alpha+\beta=9) and (\alpha\beta=20). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=729-540=189).

For real and equal roots, the discriminant must be zero. Here, \(a=1, b=-4, c=k\), so \(D=b^2-4ac=16-4k\). Thus, \(16-4k=0\), giving \(k=4\). The value 2 may result from an incorrect calculation of the discriminant equation. Exam tip: Set \(D=0\) whenever a quadratic equation has equal roots.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=k\), \(b=-6\), and \(c=9\), so \(D=(-6)^2-4(k)(9)=36-36k\). Setting this equal to zero gives \(k=1\). Exam tip: equal roots in a quadratic equation always require \(D=0\).

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(D=2^2-4(1)(5)=4-20=-16<0\), so the equation has no real roots. Therefore, the correct answer is 0. One or two real roots are possible only when the discriminant is 0 or positive, respectively. Exam tip: Check the sign of the discriminant first.

For a monic quadratic x² − Sx + P = 0, the sum of the roots is S and their product is P. Taking the proposed roots as 1 and a, their sum is 1 + a = a + 1, while their product is 1 × a = a. These values agree exactly with the coefficient pattern x² − (a + 1)x + a = 0. Therefore option A correctly explains the statement. Option B interchanges the sum and product. The roots are equal only for the special value a = 1, not for every a, so option C is not generally valid. The discriminant is (a − 1)², which is never negative, contradicting option D.

Here (\alpha+\beta=6) and (\alpha\beta=8). ((\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=15).

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha\beta=-10) and (\alpha+\beta=3), so the value is (-30).

Here (\alpha+\beta=\frac{5}{2}) and (\alpha\beta=1). (\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=\frac{17}{4}).

For equal roots, the roots are (-4) and (-4) because the product is (16). The sum is (-8), so (-p=-8) gives (p=8).

Putting (x=1) gives (1+2a=0), so (a=-\frac{1}{2}). The product is (a), so the other root is (-\frac{1}{2}).

For x² − 5x + q = 0, Vieta’s relations give the sum of the roots as 5. Let the roots be r and s, with r − s = 1. Solving r + s = 5 and r − s = 1 gives r = 3 and s = 2. Their product is q, so q = rs = 3 × 2 = 6. Therefore option A is correct. The other choices do not satisfy both the required sum and difference.

Here the sum of roots is (S) and product is (P). Therefore (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=S^2-2P).

The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is (x^2-6x-7=0).

Factoring the equation as \((x-6)(x+2)=0\) gives the roots 6 and -2. After subtracting 1 from each root, the new roots are 5 and -3, so their product is \(5\times(-3)=-15\). In such questions, finding the original roots first is a direct and reliable method.