Roots of a Quadratic Equation

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.

The sum of the roots is \((4+\sqrt{7})+(4-\sqrt{7})=8\). Comparing with the standard form \(x^2-(\text{sum of roots})x+\text{product of roots}=0\), we get \(a=-8\). Their product is \((4+\sqrt{7})(4-\sqrt{7})=16-7=9\), so \(b=9\). Therefore, \(a+b=-8+9=1\). Exam tip: In a monic quadratic, the coefficient of \(x\) is the negative of the sum of the roots, while the constant term is their product.

Since (\alpha) is a root, (\alpha^2=5\alpha+3), and (\alpha\beta=-3). The expression becomes (5\alpha+3+5\beta-3=5(\alpha+\beta)=25).

The equation can be rewritten as \((x-t)^2-49=0\), or \((x-t)^2=49\). Therefore, its roots are \(x=t+7\) and \(x=t-7\). Their positive difference is \((t+7)-(t-7)=14\). The value 7 in option A is the distance of each root from \(t\), not the difference between the roots. Exam tip: for an equation in the form \((x-a)^2=b^2\), the difference between its roots is \(2b\).

The sum (6) is positive and (c>0) is needed for both positive roots. For real roots, (36-4c\ge0), so (0<c\le9).

We use ((\alpha-5)(\beta-5)=\alpha\beta-5(\alpha+\beta)+25). Since (\alpha+\beta=4) and (\alpha\beta=-12), the value is (-7).

The sum of these roots is (\frac{5t+3}{4}), and the product is (\frac{t(t+3)}{4}). These match (-\frac{b}{a}) and (\frac{c}{a}) of the given equation.

By Vieta’s relations, the product of the roots is the constant term divided by the coefficient of \(x^2\), which is \(9m\). If the roots are \(9\) and \(r\), then \(9r=9m\), giving \(r=m\). Option C is the sum of the roots, not the other root. Exam tip: Identify the sum and product of roots before solving parameter-based quadratic questions.

The equation can be factorised as \((x-a)(x-a-4)=0\), whose expansion is \(x^2-2(a+2)x+a^2+4a\). Hence, \(x=a\) or \(x=a+4\), so option A is correct. Option C has the correct sum of roots, but its product is \(a^2-4\), not the required \(a^2+4a\). Exam tip: verify proposed roots using their sum, \(2(a+2)\), and product, \(a(a+4)\).

We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha^2+\beta^2=169-72=97) and (\alpha\beta=36), so the value is (\frac{97}{36}).

The product of the two roots is (1), so the product condition is satisfied. For real roots, the discriminant (s^2-4\ge0), so (s^2\ge4).

The sum of roots is (\frac{1}{4+\sqrt{3}}+\frac{1}{4-\sqrt{3}}=\frac{8}{13}). In (x^2+ax+b=0), the sum is (-a), so (a=-\frac{8}{13}).

For a quadratic equation \(Ax^2+Bx+C=0\), the product of its roots is \(C/A\). Here, \(A=1\) and \(C=a^2-16\), so the product of the roots is \(a^2-16\). Setting it equal to zero gives \(a^2-16=0\), or \(a^2=16\), hence \(a=4\) or \(a=-4\). Therefore, option A is correct. In exams, remember that the sum and product of roots are \(-B/A\) and \(C/A\), respectively.

Let the roots be \(\alpha,\beta\). Reciprocal roots must satisfy \(\alpha\beta=1\). By Vieta’s formula, \(\alpha\beta=c/a\), so \(c/a=1\Rightarrow a=c\). The condition \(b=0\) only makes the sum of roots zero. Exam tip: check the product first for reciprocal roots.

Let the roots be \(5r\) and \(3r\). Their sum is \(5r+3r=16\), so \(r=2\). Hence, their product is \((5r)(3r)=15r^2=15\times4=60\). In the monic quadratic equation \(x^2-16x+q=0\), the product of the roots equals the constant term \(q\). Therefore, \(q=60\). Exam tip: For \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\).

Factoring the quadratic gives \(x^2-(u+2v)x+2uv=(x-u)(x-2v)\). Hence, \(x=u\) or \(x=2v\), so option A is correct. In option B, the sum of the proposed roots is \(2u+v\), which generally does not equal the given sum \(u+2v\). Exam tip: verify roots using their sum \(u+2v\) and product \(2uv\).

We use (\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha+\beta=7), so (7r=84) and (r=12).