Methods of Solving Quadratic Equations

The first equation has roots (\frac{1}{2},\frac{5}{2}), and the second has roots (\frac{3}{2},\frac{4}{3}), so none of the listed values is common. In exams, solve both equations correctly before comparing.

The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.

The other root is (9), so (q=6\times9=54). In exams, when (a=1), (c) equals the product of roots.

(4x^2+4x-3=(2x-1)(2x+3)), so (x=\frac{1}{2},-\frac{3}{2}) is correct. In exams, change signs carefully from factors.

Since (7=(\sqrt{7})^2) and the middle term is (2\sqrt{7}x), it is ((x+\sqrt{7})^2). In exams, identify perfect squares even with irrational coefficients.

((x+\sqrt{7})^2=0), so the repeated root is (-\sqrt{7}). In exams, ((x+a)^2=0) gives (x=-a).

The equation is equivalent to ((x-u)(x-v)=0), so the roots are (u) and (v). In exams, apply the same rule to symbolic factors.

It is ((x-m)^2-n^2=0), so (x-m=\pm n) and (x=m\pm n). In exams, recognize the difference of squares.

Dividing the whole equation by (16) gives (x^2-(a+b)x+ab=0). In exams, removing the common factor first makes solving easier.

Since (x^4=(x^2)^2=y^2), the new equation is (y^2-13y+36=0). In exams, substitution simplifies a difficult form.

From (y^2-13y+36=0), (y=4,9), so (x^2=4,9) and (x=\pm2,\pm3). In exams, do not forget to return to (x).

Multiplying both sides by (4x) gives (4+4x^2=17x), that is (4x^2-17x+4=0). In exams, remember the condition (x\neq0).

(4x^2-17x+4=(4x-1)(x-4)), so (x=\frac{1}{4}) and (4). In exams, check whether obtained roots are valid in the original equation.

Cross multiplication gives ((x+3)^2=16x), so (x^2+6x+9-16x=0), and (x^2-10x+9=0). In exams, cross multiply carefully.

(x^2-10x+9=(x-1)(x-9)), so (x=1) and (x=9). In exams, check solutions against excluded denominator values.

(D=(-8)^2-4(1)(3)=52), so (x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}). In exams, simplify the square root.

The sum of roots is (-\frac{p}{5}), so (-\frac{p}{5}=-7) gives (p=35). In exams, remember the sum formula (-\frac{b}{a}).

The product of roots is (\frac{p}{6}), so (\frac{p}{6}=\frac{1}{2}) gives (p=3). In exams, use the product formula (\frac{c}{a}).

(\alpha+\beta=17) and (\alpha\beta=70), so (\alpha^2+\beta^2=17^2-2(70)=149). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}). In exams, first write sum and product in reciprocal questions.