Roots of a Quadratic Equation

Here (\alpha+\beta=\frac{11}{3}) and ((\alpha-\beta)^2=\frac{25}{9}). Using ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta), we get (\alpha\beta=\frac{8}{3}), so (p=8).

For equal roots, put (D=0). Simplifying (4(k+2)^2-4(k^2+5)=0) gives (4k-1=0), so (k=\frac{1}{4}).

Let the roots be (r) and (2r), then (2r^2=4) gives (r=\pm\sqrt{2}). Since (3r=-\frac{m}{2}), we get (m=\pm6\sqrt{2}).

We use (\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}). Since (\alpha^2+\beta^2=21) and ((\alpha\beta)^2=4), the value is (\frac{21}{4}).

For reciprocal roots, (\frac{k}{k}=1), so (k\neq0) is needed. For real roots, (64-4k^2\ge0), hence (k^2\le16).

By Vieta’s formulas, the product of the roots is 12, so (r(r+1)=12), or (r^2+r-12=0). Thus, r=3 or r=-4. The sum of the roots is therefore 7 or -7. Since r+(r+1)=-p, we obtain p=-7 or p=7; hence, option A is correct. Exam tip: for x^2+px+q=0, the sum of the roots is -p and their product is q.

A quadratic equation has real roots only when its discriminant satisfies \(D\ge0\). Here, \(A=4\), \(B=-4(a-1)\), and \(C=a^2-4a\). Thus, \(D=B^2-4AC=16(a-1)^2-16(a^2-4a)=16(2a+1)\). Therefore, \(16(2a+1)\ge0\), giving \(a\ge-\frac12\). Hence, option A is correct; the condition \(a\le1\) in option B does not follow from the discriminant. Exam tip: simplify the discriminant completely before solving the parameter inequality.

The roots are (3) and (5). Substitution gives (\frac{5}{1}+\frac{7}{3}=\frac{22}{3}), so none of the options is correct; the correct value should be (\frac{22}{3}).

One root can be (\alpha=2), which makes (\alpha-2=0). Therefore the expression is undefined; always check zero denominators first in exams.

Here (\alpha+\beta=\frac{5}{2}) and (\alpha\beta=\frac{3}{2}). Using (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)), we get (\frac{35}{8}).

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=m-5\), and \(c=9\), so \(D=b^2-4ac=(m-5)^2-36=0\). Hence, \(m-5=\pm6\), giving \(m=11\) or \(m=-1\). Also, the product of the roots is \(c/a=9>0\), so the roots cannot have opposite signs. Exam tip: For equal roots of a quadratic equation, directly use \(b^2-4ac=0\).

For a quadratic equation \(x^2+ax+b=0\), the sum of the roots is \(-a\) and their product is \(b\). Here, the sum is \((2+\sqrt{3})+(2-\sqrt{3})=4\), so \(a=-4\). Their product is \((2+\sqrt{3})(2-\sqrt{3})=4-3=1\), so \(b=1\). Therefore, \(a+b=-4+1=-3\). Exam tip: Use Vieta’s formulas directly to find the sum and product of the roots.

Since (\alpha^2=3\alpha+1), the expression becomes (3\alpha+1+3\beta-1=3(\alpha+\beta)=9), so no option is correct. The correct value is (9).

Use (\alpha^2=3\alpha+1) and (\alpha\beta=-1). Then the expression is (3\alpha+1+3\beta-1=3(\alpha+\beta)=9).

The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, (16-4c\ge0), hence (0<c\le4).

We use ((\alpha+3)(\beta+3)=\alpha\beta+3(\alpha+\beta)+9). Since (\alpha+\beta=2) and (\alpha\beta=-8), the value is (7).

The sum is (\frac{3t+1}{2}) and the product is (\frac{t(t+1)}{2}). These match (-\frac{b}{a}) and (\frac{c}{a}) for every (t).

Putting (x=3) gives (9-3(m-2)+m-6=0), so (m=\frac{9}{2}). The product is (-\frac{3}{2}), so the other root is (-\frac{1}{2}); hence no option is correct.

Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).

We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha^2+\beta^2=45) and (\alpha\beta=18), so the value is (\frac{5}{2}).

We would need (\sin \theta+\cos \theta=4), but its maximum is (\sqrt{2}). Therefore no real (\theta) is possible.

By Vieta’s formula, the product of the roots of \(x^2-2x+k=0\) is \(k\). Here, the product is \(\tan\theta\cdot\cot\theta=1\), so \(k=1\). The value 2 is related to the sum of the roots, not their product. Exam tip: for a monic quadratic \(x^2+bx+c=0\), the product of the roots is \(c\).

Since \((2+\sqrt{3})(2-\sqrt{3})=1\), the given reciprocals are \(2-\sqrt{3}\) and \(2+\sqrt{3}\). Their sum is \(4\). For \(x^2+ax+b=0\), the sum of the roots is \(-a\). Thus, \(-a=4\), giving \(a=-4\). Exam tip: For a quadratic equation, use the sum-of-roots relation \(-\frac{\text{coefficient of }x}{\text{coefficient of }x^2}\).

By Vieta’s formula, the product of the roots of \(Ax^2+Bx+C=0\) is \(C/A\). Here, \(A=1\) and \(C=a^2-1\), so the product of the roots is \(a^2-1\). Setting it equal to zero gives \(a^2-1=0\), or \(a^2=1\), hence \(a=1\) or \(a=-1\). Therefore, option A is correct. Exam tip: Use Vieta’s formulas directly when a question asks for the sum or product of roots, and remember to consider both values obtained after solving the parameter equation.

By Vieta’s formulas, (\alpha+\beta=6) and (\alpha\beta=n). Using (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta), we get (20=6^2-2n=36-2n). Hence, (2n=16) and (n=8). Exam tip: For the sum of squares of roots, use ((\alpha+\beta)^2-2\alpha\beta) rather than finding the roots individually.