Methods of Solving Quadratic Equations

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}). In exams, first write sum and product in reciprocal questions.

The sum of roots is (-\frac{b}{a}=-\frac{-11}{3}=\frac{11}{3}). In exams, keep the sign of (b) carefully.

The product of roots is (\frac{c}{a}=\frac{6}{3}=2). In exams, use (\frac{c}{a}) for the product.

\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{11}{3}\right)^2-8=\frac{49}{9}\). In exams, use this identity for the square of difference.

The direct answer is option A: \(n>16\). For a quadratic equation to have no real roots, its discriminant must be negative: \(D<0\). In \(x^2-8x+n=0\), we have \(a=1\), \(b=-8\), and \(c=n\). Therefore \(D=(-8)^2-4(1)(n)=64-4n\). Apply the condition: \(64-4n<0\). Subtract 64: \(-4n<-64\). Dividing by \(-4\) reverses the inequality, giving \(n>16\). Option A is correct. Option B, \(n<16\), makes the discriminant positive and gives two real roots. Option C, \(n=16\), makes \(D=0\), giving one repeated real root, not no real roots. Option D, \(n\le16\), includes values that produce real roots, so it is too broad. Remember the discriminant test: \(D>0\) means two real roots, \(D=0\) means equal real roots, and \(D<0\) means no real roots.

For two distinct real roots, (D>0), so (64-4n>0) and (n<16). In exams, connect (D>0) with distinct real roots.

Expanding option A gives (3x+1)(x−2) = 3x² − 6x + x − 2 = 3x² − 5x − 2, so it is the correct factorised form. Option B gives a middle term of +5x, option C gives x, and option D gives −7x. In an exam, expand the factors and compare the result with the original quadratic equation.

The factorisation is \(3x^2-5x-2=(3x+1)(x-2)\). By the zero-product rule, \(3x+1=0\) or \(x-2=0\), giving the roots \(x=-\frac{1}{3}\) and \(x=2\). Option B has the signs of both roots reversed. As an exam check, the sum of the roots should be \(\frac{5}{3}\) and their product should be \(-\frac{2}{3}\).

Rewrite the equation by completing the square: \(x^2+6x+2=0\Rightarrow (x+3)^2=7\). Thus, \(x+3=\pm\sqrt{7}\), giving \(x=-3\pm\sqrt{7}\). Option B has the wrong sign before 3, so it does not satisfy the equation. In an exam, remember that \(\pm\) represents both roots.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=4) and (\alpha\beta=-12), so the value is (-48). In exams, factor the expression first.

Let the roots be (-r) and (-2r), then (2r^2=16) gives (r=2\sqrt{2}), and (p=3r=6\sqrt{2}). In exams, keep signs of both roots carefully.

Here (D=(-5)^2-4(1)(2)=17), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{17}). In exams, the difference of roots can be found directly from (D).

Here (D=(-6)^2-4(3)(7)=-48<0), so there are no real roots. In exams, (D<0) means no real roots.

(3x^2-6x+7=3(x-1)^2+4), so it cannot be zero for real (x). In exams, completed square form also shows the nature of real roots.

The roots are (2,6), so new roots are (4,8), and the equation is ((x-4)(x-8)=0). In exams, form the new roots and then the new equation.

(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}), where (\alpha+\beta=10) and (\alpha\beta=21), so the value is (\frac{100-42}{21}=\frac{58}{21}). In exams, convert expressions into sum and product.

For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-10\), and \(c=25\), so \(D=(-10)^2-4(k)(25)=100-100k\). Thus, \(100-100k=0\), giving \(k=1\). Exam tip: identify the coefficient of \(x^2\) as \(a=k\) before applying the discriminant formula.

(D=4(k+3)^2-4k^2=0) gives ((k+3)^2=k^2), so (6k+9=0) and (k=-\frac{3}{2}). In exams, expand squares carefully.

((x-3)(x-7)=x^2-10x+21), so (x^2-10x+21=10) gives (x^2-10x+11=0). In exams, bring all terms to one side after expansion.

Here (D=(-10)^2-4(1)(11)=56), so (x=\frac{10\pm2\sqrt{14}}{2}=5\pm\sqrt{14}). In exams, simplify (D) correctly.

First \(x^2+5x+\frac{3}{2}=0\) is obtained, then adding \(\frac{25}{4}\) gives \(\left(x+\frac{5}{2}\right)^2=\frac{19}{4}\). In exams, divide by (a) first when \(a\neq1\).

(10x^2-13x+3=(10x-3)(x-1)), so the roots are (1) and (\frac{3}{10}). In exams, set each linear factor equal to zero.

Here (ac=180) and (-15+(-12)=-27), so the correct split is (-15x-12x). In exams, match both sum (b) and product (ac).

(18x^2-27x+10=(3x-2)(6x-5)), so the roots are (\frac{2}{3}) and (\frac{5}{6}). In exams, write fractional roots in simplest form.

For equal roots, (D=0), so (4(k-2)^2-4(k^2-9)=0) gives (-4k+13=0). In exams, expand (D) carefully in parameter questions.