Roots of a Quadratic Equation

The product of roots is (\frac{m-1}{m-1}=1), so (m\neq1) is needed. For real roots, (D=16m\ge0), hence (m\ge0) and (m\neq1).

The equation becomes ((x-(a+1))(x-(a+5))=0). So the roots are (a+1) and (a+5), hence the positive difference is (4).

Use the product-of-roots relation for ax^2+bx+c=0: the product of the roots is c/a. Here a=2 and c=p(p+2), so the product is p(p+2)/2. If one root is p, let the other be r. Then p·r=p(p+2)/2. For p≠0, cancellation gives r=(p+2)/2. The same result is also seen by factoring the polynomial as (x−p)(2x−p−2)=0. Thus the roots are p and (p+2)/2, so option A is correct. The proposed alternatives fail the product-of-roots test.

Here (\alpha+\beta=\frac{16}{5}) and (\alpha\beta=\frac{p}{5}). Using ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta), we get (p=11).

For equal roots, put (D=0). From (4(a+5)^2-4(a^2+18)=0), we get (10a+7=0), so (a=-\frac{7}{10}).

Factor the original quadratic: x^2 - 7x + 12 = (x - 3)(x - 4), so alpha and beta are 3 and 4. Applying the stated transformation to each root gives 2alpha + 3 = 2(3) + 3 = 9 and 2beta + 3 = 2(4) + 3 = 11. We now need the quadratic whose roots are 9 and 11.

For roots r and s, the equation is x^2 - (r+s)x + rs = 0. Their sum is 9 + 11 = 20 and their product is 9 × 11 = 99. Therefore the required equation is x^2 - 20x + 99 = 0, so option A is correct. The coefficient -14 in option B does not use the transformed-root sum, and the other choices have an incorrect sign or constant term.

Let the roots be (2r) and (5r). From (10r^2=\frac{10}{3}), (r=\pm\frac{1}{\sqrt{3}}), so (7r=-\frac{m}{3}) gives (m=\pm7\sqrt{3}).

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.

Here (\alpha^2+\beta^2=64-4=60) and ((\alpha\beta)^2=4). Thus (\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{60}{4}=15).

The product of roots is (\frac{k}{k}=1), so (k\neq0) is needed. For real roots, (144-4k^2\ge0), hence (k^2\le36).

From (r(r+2)=24), we get (r=4) or (r=-6). The sum of roots is (10) or (-10), so (p=-10) or (p=10).

For real roots, (D\ge0) is required. Here (D=64(a-2)^2-64(a^2-6a)=128(a+2)), so (a\ge-2); hence none of these options is exact.

For real roots, (D\ge0) is required. Here (D=36(a-1)^2-36(a^2-4a-5)=72a+216), so the exact condition is (a\ge-3), not (a\ge-\frac{7}{2}).

Here, \(A=9\), \(B=-6(a-1)\), and \(C=a^2-4a-5\). For real roots, the discriminant must satisfy \(D=B^2-4AC\ge0\). Thus, \(D=36(a-1)^2-36(a^2-4a-5)=72(a+3)\), so \(72(a+3)\ge0\), giving \(a\ge-3\). Option B reverses the required inequality and is therefore not the general condition for real roots. Exam tip: For a quadratic to have real roots, begin by applying \(D\ge0\).

The roots are (4) and (5). Direct substitution gives (\frac{6}{2}+\frac{7}{3}=\frac{16}{3}), so option (A) should be correct.

The roots are (5) and (6). Direct substitution gives (\frac{6}{4}+\frac{7}{5}=\frac{29}{10}), so none of the given options is correct.

The roots are (5) and (6). Hence (\frac{6}{4}+\frac{7}{5}=\frac{15}{10}+\frac{14}{10}=\frac{29}{10}).

Here (\alpha+\beta=3) and (\alpha\beta=\frac{5}{4}). Using (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)), we get (\frac{63}{4}), so none of the options is correct.

Here (\alpha+\beta=3) and (\alpha\beta=\frac{5}{4}). Thus (\alpha^3+\beta^3=27-\frac{45}{4}=\frac{63}{4}).

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=m-2\), and \(c=25\), so \((m-2)^2-100=0\). Therefore, \(m-2=\pm10\), giving \(m=12\) or \(m=-8\). Exam tip: For equal-root questions, immediately set the discriminant equal to zero.