Introduction to Quadratic Equations

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\). Here, \(a=1, b=-4, c=1\), so \(\alpha+\beta=4\) and \(\alpha\beta=1\). Therefore, \(\alpha+\beta+\alpha\beta=4+1=5\). Exam tip: Use the sum and product of roots directly instead of solving the quadratic equation.

For the quadratic equation \\(x^2+px+q=0\\), the product of the roots equals the constant term \\(q\\). Since the roots are 1 and \\(p\\), \\(q=1\cdot p=p\\). The sum condition also gives \\(1+p=-p\\), so \\(p=-\\frac{1}{2}\\); nevertheless, the required relation remains \\(q=p\\). Exam tip: In \\(x^2+bx+c=0\\), the product of the roots is always \\(c\\).

Bringing \(x(x-3)=4\) to one side gives \(x^2-3x-4=0\), where the coefficient of \(x^2\) is 1. Hence it is quadratic. Option D is an identity because both sides are identical. Exam tip: check that the highest power is 2.

The product of roots is (\frac{c}{a}). In the first option, (\frac{-5}{2}<0), so the product is negative.

The area of a right triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Hence, \(\frac{1}{2}(x+1)(x+3)=30\), giving \((x+1)(x+3)=60\). Expanding, \(x^2+4x+3=60\), so the quadratic equation is \(x^2+4x-57=0\). Option D results from mishandling the factor \(\frac{1}{2}\). Exam tip: For triangle-area problems, include the factor \(\frac{1}{2}\) before expanding the algebraic expression.

By Vieta’s formula, the sum of the roots of \(ax^2+bx+c=0\) is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=1\) and \(b=2\), so \(\alpha+\beta=-2\). Therefore, \((\alpha+\beta)^2=(-2)^2=4\). Exam tip: square the complete sum, including its sign; the negative sign disappears only after squaring.

A quadratic equation has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=4, c=k\), so \(D=4^2-4(1)(k)=16-4k\). Thus, \(16-4k<0\), which gives \(k>4\). At \(k=4\), the discriminant is zero and the equation has one repeated real root, so that option is incorrect. Exam tip: for ‘no real roots,’ always use the condition \(D<0\).

The equation can be rewritten as \((x-a)^2-9=0\). Thus, \((x-a)^2=9\), giving the roots \(x=a+3\) and \(x=a-3\). The difference between the larger and smaller roots is \((a+3)-(a-3)=6\). Option D represents only one root, while option C is not generally the difference between the roots. Exam tip: For this form, complete the square and identify the two roots directly.

Here ((6x+1)(x-4)=6x^2-23x-4), and subtracting (5x) gives (6x^2-28x-4=0). First expand and then bring all terms to one side.

A quadratic equation in x must have a non-zero coefficient of x². In the given equation, that coefficient is n² − 16. Therefore, the necessary condition is n² − 16 ≠ 0. Solving the equality that must be avoided gives n² = 16, so n = 4 or n = −4. Hence both values must be excluded, which can be written compactly as n ≠ ±4. Option A excludes only n = 4 and still permits −4, while option B excludes only −4 and still permits 4; both are incomplete. If option D were used, the x² coefficient would become zero and the equation would reduce to −3x + 7 = 0, which is linear. Thus option C is the only complete and correct answer.

Expand the squares: \\( (2x-3)^2=4x^2-12x+9 \\) and \\( (x+5)^2=x^2+10x+25 \\). This gives \\(5x^2-2x+34=34\\); subtracting 34 from both sides yields \\(5x^2-2x=0\\). This matches the standard quadratic form \\(ax^2+bx+c=0\\), with \\(c=0\\). Exam tip: after expansion, bring all terms to one side and equate the result to zero.

Multiplying the whole equation by (6) gives (2x^2+2-3x+6=30). Therefore the standard form is (2x^2-3x-22=0).

A given root must make the equation equal to zero. Substituting (x=-4) gives (2(-4)^2+p(-4)-12=0), or (32-4p-12=0). Thus (20-4p=0), so (p=5). Option B results from a sign error and gives (p=-5). Exam tip: Substitute the given root directly into the quadratic equation to find the unknown parameter.

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 \(\left(-\frac{8}{3}\right)^2-2\cdot\frac{4}{3}=\frac{40}{9}\).