Methods of Solving Quadratic Equations

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.

The governing concept is the zero-product rule: if the product of two real expressions is zero, at least one factor must be zero. Therefore, from (x−3)(x−7)=0, set the factors separately equal to zero: x−3=0 gives x=3, and x−7=0 gives x=7. Thus the solution set is {3,7}, so option A is correct. Option B incorrectly changes both signs, while options C and D do not make either original factor zero in the required alternatives. Substitution confirms both answers: for x=3 the first factor is zero, and for x=7 the second factor is zero.

((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.

The sum of roots is (3), and (\frac{p-1}{4}=3), so (p=13). In exams, use (-\frac{b}{a}) for the sum of roots.

First \(x^2-\frac{22}{7}x+1=0\) is obtained, then \(\left(x-\frac{11}{7}\right)^2=\frac{72}{49}\). In exams, divide by (a) first when \(a\neq1\).

Since \(\left(x-\frac{11}{7}\right)^2=\frac{72}{49}\), \(x=\frac{11\pm6\sqrt{2}}{7}\). In exams, simplify \(\sqrt{72}=6\sqrt{2}\).

For real and equal roots, (D=0), so (324-4m=0) gives (m=81). In exams, equal roots mean (D=0).

The first equation has roots (2,\frac{6}{5}), and the second has roots (\frac{3}{2},\frac{4}{3}), so there is no common root among the given values. In exams, solve both equations before comparing.