Methods of Solving Quadratic Equations

Let the roots be (-r) and (-2r), then (2r^2=36) gives (r=3\sqrt{2}), and (p=3r=9\sqrt{2}). In exams, keep signs of both roots carefully.

Here (D=(-9)^2-4(1)(5)=61), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{61}). In exams, the difference of roots can be found directly from (D).

Here (D=(-10)^2-4(5)(13)=-160<0), so there are no real roots. In exams, (D<0) means no real roots.

(5x^2-10x+13=5(x-1)^2+8), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.

The roots are (2,10), so new roots are (6,14), and the equation is ((x-6)(x-14)=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=18) and (\alpha\beta=77), so the value is (\frac{324-154}{77}=\frac{170}{77}). In exams, convert expressions into sum and product.

A quadratic equation \(ax^2+bx+c=0\) has equal roots when its discriminant \(b^2-4ac\) is zero. Here, \(a=k\), \(b=-18\), and \(c=81\). Thus, \((-18)^2-4(k)(81)=0\), giving \(324-324k=0\) and hence \(k=1\). Exam tip: identify the coefficient of \(x^2\) as \(a\) before applying the discriminant formula.

(D=4(k+5)^2-4k^2=0) gives ((k+5)^2=k^2), so (10k+25=0) and (k=-\frac{5}{2}). In exams, expand squares carefully.

((x-5)=0) or ((x-11)=0), so (x=5) or (x=11). In exams, set each factor equal to zero separately.

((x-5)(x-11)=x^2-16x+55), so (x^2-16x+55=18) gives (x^2-16x+37=0). In exams, bring all terms to one side after expansion.

Here (D=(-16)^2-4(1)(37)=108), so (x=\frac{16\pm6\sqrt{3}}{2}=8\pm3\sqrt{3}). In exams, simplify (D) correctly.

(16x^2-38x+15=(8x-3)(2x-5)), so the roots are (\frac{3}{8}) and (\frac{5}{2}). In exams, do not invert fractional roots.

Here (ac=600) and (-30+(-20)=-50), so the correct split is (-30x-20x). In exams, even for large (ac), match both sum and product.

(24x^2-50x+25=(6x-5)(4x-5)), so the roots are (\frac{5}{6}) and (\frac{5}{4}). In exams, keep the denominator coefficients correctly.

For equal roots, (D=0), so ((k-4)^2=k^2-25) and (k=\frac{41}{8}). In exams, handle constant terms carefully while expanding squares.

The sum of roots is (4), and (\frac{p+5}{6}=4), so (p=19). In exams, use (-\frac{b}{a}) for the sum.

First (x^2-4x+\frac{7}{11}=0) is obtained, then ((x-2)^2=\frac{37}{11}). In exams, divide by (a) first when (a\neq1).

Since ((x-2)^2=\frac{37}{11}), (x=2\pm\sqrt{\frac{37}{11}}=2\pm\frac{\sqrt{407}}{11}). In exams, rationalize the denominator.

For real and equal roots, (D=0), so (484-4m=0) gives (m=121). In exams, equal roots indicate (D=0).

The first equation has roots (\frac{3}{2},\frac{9}{4}), and the second has roots (\frac{3}{2},\frac{10}{9}). In exams, solve both equations separately for the common root.