Methods of Solving Quadratic Equations

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=12) and (\alpha\beta=-28), so the value is (-336). In exams, factor the expression first.

Let the roots be (-r) and (-2r), then (2r^2=64) gives (r=4\sqrt{2}), and (p=3r=12\sqrt{2}). In exams, keep signs of both roots carefully.

Here (D=(-13)^2-4(1)(7)=141), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{141}). In exams, the difference of roots can be found directly from (D).

Here (D=(-14)^2-4(7)(19)=-336<0), so there are no real roots. In exams, (D<0) means no real roots.

(7x^2-14x+19=7(x-1)^2+12), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.

The roots are (4,12), so new roots are (10,18), and the equation is ((x-10)(x-18)=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=26) and (\alpha\beta=165), so the value is (\frac{676-330}{165}=\frac{346}{165}). In exams, convert expressions into sum and product.

For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-26\), and \(c=169\). Thus, \(D=(-26)^2-4(k)(169)=676-676k=0\), giving \(k=1\). Values such as \(13\) or \(169\) do not make the discriminant zero. Exam tip: For equal roots in a parameter-based quadratic equation, immediately apply \(b^2-4ac=0\).

(D=4(k+7)^2-4k^2=0) gives ((k+7)^2=k^2), so (14k+49=0) and (k=-\frac{7}{2}). In exams, expand squares carefully.

((x-7)(x-15)=x^2-22x+105), so (x^2-22x+105=26) gives (x^2-22x+79=0). In exams, bring all terms to one side after expansion.

Here (D=(-22)^2-4(1)(79)=168), so (x=\frac{22\pm2\sqrt{42}}{2}=11\pm\sqrt{42}). In exams, simplify (D) correctly.