Introduction to Quadratic Equations

A monic equation is (x^2-(\text{sum})x+\text{product}=0). Substituting sum (-9) gives (x^2+9x+20=0).

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=-2k\), and \(c=25\), so \(D=b^2-4ac=(-2k)^2-4(1)(25)=4k^2-100\). Setting \(D=0\) gives \(4k^2-100=0\), hence \(k^2=25\) and \(k=\pm5\). Therefore, option A is correct. Choosing only \(k=5\) or only \(k=-5\) is incomplete because both values are possible. Exam tip: For equal roots, immediately use \(D=0\).

For a quadratic equation \(ax^2+bx+c=0\) to have real roots, its discriminant must satisfy \(D=b^2-4ac\geq0\). Here, \(a=1, b=2k, c=16\), so \(D=(2k)^2-4(1)(16)=4k^2-64\). Thus, \(4k^2-64\geq0\Rightarrow k^2\\geq16\), which gives \(k\leq-4\) or \(k\geq4\). In option B, \(k\) lies between \(-4\) and \(4\), making the discriminant negative. Exam tip: Whenever real roots are asked for, begin with the condition \(D\geq0\).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here it is (\frac{\frac{10}{3}}{\frac{7}{3}}=\frac{10}{7}).

For the quadratic equation \(x^2-sx+p=0\), the sum of the roots is \(s\), and their product is \(p\). Thus, \(s=4+5=9\) and \(p=4\times5=20\). Therefore, \(s+p=9+20=29\), so option A is correct. Exam tip: compare the equation directly with the standard form \(x^2-(\text{sum of roots})x+(\text{product of roots})=0\).

For a quadratic equation to have real and distinct roots, its discriminant \(D=b^2-4ac\) must be positive. Here, \(a=1, b=-8, c=k\), so \(D=(-8)^2-4(1)(k)=64-4k\). Thus, \(64-4k>0\), which gives \(k<16\). When \(k=16\), the roots are equal, so option B is incorrect. Exam tip: use \(D>0\) specifically for distinct real roots.

If the roots are \(\alpha,\beta\), then \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\). Here \(\left(-\frac{8}{3}\right)^2-2\cdot\frac{4}{3}=\frac{40}{9}\).

By Vieta’s formulas, \(\alpha+\beta=-\frac{7}{1}=-7\) and \(\alpha\beta=\frac{10}{1}=10\). Therefore, \((\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=10+2(-7)+4=0\). Option B is only the value of \(\alpha\beta\), not of the complete expression. Exam tip: For expressions involving roots, first find the sum and product of the roots using Vieta’s formulas.

Expand the two products: (x+2)(x+5)=x^2+7x+10) and (x-1)(x+4)=x^2+3x-4). Their sum is 2x^2+10x+6). Bringing 60 to the left gives 2x^2+10x+6-60=0), which simplifies to 2x^2+10x-54=0). Option B stops before subtracting 60, while option C does not represent the correctly simplified equation. Exam tip: In standard form, collect all terms on one side and write zero on the other side.

For this quadratic equation, \(a=1\), \(b=-6\), and \(c=11\). Its discriminant is \(D=b^2-4ac=(-6)^2-4(1)(11)=36-44=-8\). Since \(D<0\), the equation has no real roots. Exam tip: \(D=0\) gives equal real roots, while \(D>0\) gives two distinct 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 \(4m\). Since one root is \(4\), the other root is \(\frac{4m}{4}=m\). Remember that \(m+4\) is the sum of the roots, not the other root.

Here, the discriminant is \(D=b^2-4ac=(-9)^2-4(2)(10)=81-80=1\). The difference between the roots is \(\frac{\sqrt{D}}{|a|}\), so it equals \(\frac{\sqrt{1}}{2}=\frac{1}{2}\). Alternatively, \(2x^2-9x+10=(2x-5)(x-2)\), giving roots \(\frac{5}{2}\) and \(2\), whose difference is \(\frac{1}{2}\). Exam tip: use \(\frac{\sqrt{D}}{|a|}\) when only the difference between roots is required.

For the quadratic to be a perfect square, its left-hand polynomial must be expressible as \(x^2-14x+c=(x-7)^2\). Since \((x-7)^2=x^2-14x+49\), we get \(c=49\). Exam tip: for \(x^2+bx+c\) to be a perfect square, \(c=\left(\frac{b}{2}\right)^2\); here, \(b=-14\).

The sum of the roots is \((-7)+(-7)=-14\). For a quadratic equation \(x^2+bx+c=0\), the sum of the roots is \(-b\). Thus, \(-b=-14\), giving \(b=14\). Option B is the sum of the roots, not the value of the coefficient \(b\). Exam tip: Use the relation \(\text{sum of roots}=-\frac{\text{coefficient of }x}{\text{coefficient of }x^2}\).

In the first option, the sum is (-\frac{b}{a}=-7) and the product is (\frac{c}{a}=12). So the sum is negative and the product is positive.

A quadratic equation has the form \(ax^2+bx+c=0\), where \(a\ne0\). In option A, the highest power of x is 2. Option B is cubic, while D is linear. Exam tip: check the degree after simplifying fractions.

If the sum of roots is (13) and product is (40), the equation is (x^2-13x+40=0). Remember the monic form formula.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the value is (\frac{7}{12}).

For option A, \(x^2+16x+64=(x+8)^2\). Thus, \((x+8)^2=0\) gives both roots as \(x=-8\), so they are equal and negative. In option B, \((x-8)^2=0\), giving equal but positive roots. Exam tip: For equal roots, first check that the discriminant is zero, then determine the sign of the root.

Expanding gives \\(x+a\\)^2 = x^2 + 2ax + a^2\\). Comparing the coefficients of x, \\(2a=-12\\), so \\(a=-6\\). The constant term gives \\(a^2=36\\), which alone allows both 6 and -6; the coefficient of x identifies -6 uniquely. Exam tip: For an identity, compare coefficients of like powers of the variable.

The equation can be rewritten as \((3x+2k)^2=0\). Thus \(3x+2k=0\), giving the repeated root \(x=-\frac{2k}{3}\). Equivalently, the discriminant is \(D=(12k)^2-4(9)(4k^2)=0\), so the roots are equal and real. Exam tip: for a quadratic equation, \(D=0\) indicates two equal real roots.

If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (25), so it is not possible.

For equal roots, (p^2-64=0) gives (p=\pm8). For equal positive roots, (-\frac{p}{2}>0), so (p=-8).

The direct answer is option A: k ≠ 1. A number is a root of an equation when substituting that number makes the left-hand side equal to zero. Put x = -1 into kx² + 3x + 2. First, (-1)² = 1, so kx² = k. Next, 3x = 3(-1) = -3. Therefore the expression becomes k - 3 + 2 = k - 1. Since -1 is not a root, this value must not be zero: k - 1 ≠ 0, hence k ≠ 1. Option A is correct because it states exactly this condition. Option B, k = 1, is wrong because then the value is 1 - 1 = 0, so -1 would actually be a root. Option C, k ≠ 3, is not the condition obtained; for example, k = 1 satisfies k ≠ 3 but still makes -1 a root. Option D, k = 3, is also unnecessarily specific and does not describe all allowed values; k = 3 is allowed, but many other values are allowed too. Exam cue: for a proposed root, substitute it and require the expression to be zero; for a non-root, require it to be nonzero.

The monic equation is (x^2-(\text{sum})x+\text{product}=0). Using sum (0) and product (-81) gives (x^2-81=0).