Methods of Solving Quadratic Equations

The product of roots is (\frac{c}{a}=\frac{12}{7}). In exams, use (\frac{c}{a}) for the product.

\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{25}{7}\right)^2-\frac{48}{7}=\frac{289}{49}\). In exams, convert fractions to a common denominator.

For no real roots, (D<0), so (196-4n<0) and (n>49). In exams, connect (D<0) with no real roots.

For two distinct real roots, (D>0), so (196-4n>0) and (n<49). In exams, connect (D>0) with distinct real roots.

Expanding option A gives \((7x+2)(x-3)=7x^2-21x+2x-6=7x^2-19x-6\), so it is the correct factorised form. Option B produces a middle term of \(+19x\), while options C and D produce \(-11x\) and \(+11x\), respectively. Exam tip: always expand the factors to verify the middle term and constant term.

The quadratic factors as \(7x^2-19x-6=(7x+2)(x-3)\). Therefore, \((7x+2)(x-3)=0\) gives \(x=-\frac{2}{7}\) or \(x=3\). In option B, the signs of both roots are incorrect. Exam tip: After factorisation, set each factor equal to zero and carefully reverse the sign when isolating the root.

Adding (36) to (x^2+12x=-8) gives ((x+6)^2=28). In exams, add the same number to both sides.

Completing the square gives \(x^2+12x+8=0\Rightarrow (x+6)^2=28\). Therefore, \(x+6=\pm\sqrt{28}=\pm2\sqrt{7}\), so the roots are \(x=-6\pm2\sqrt{7}\). Option B has the wrong sign for the constant term in the roots. Exam tip: simplify \(\sqrt{28}\) to \(2\sqrt{7}\).

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=10) and (\alpha\beta=-24), so the value is (-240). In exams, factor the expression first.

Let the roots be (-r) and (-2r), then (2r^2=49) and (p=3r=\frac{21\sqrt{2}}{2}). In exams, do not forget to rationalize the denominator.

Here (D=(-11)^2-4(1)(6)=97), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{97}). In exams, the difference of roots can be found directly from (D).

Here (D=(-12)^2-4(6)(17)=-264<0), so there are no real roots. In exams, (D<0) means no real roots.

(6x^2-12x+17=6(x-1)^2+11), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.

The roots are (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=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=22) and (\alpha\beta=117), so the value is (\frac{484-234}{117}=\frac{250}{117}). In exams, convert expressions into sum and product.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=k\), \(b=-22\), and \(c=121\), so \(D=(-22)^2-4(k)(121)=484-484k\). Setting this equal to zero gives \(k=1\). Therefore, option A is correct. Exam tip: identify \(a\), \(b\), and \(c\) carefully before applying \(D=0\).

(D=4(k+6)^2-4k^2=0) gives ((k+6)^2=k^2), so (12k+36=0) and (k=-3). In exams, expand squares carefully.

((x-6)=0) or ((x-13)=0), so (x=6) or (x=13). In exams, set each factor equal to zero separately.

((x-6)(x-13)=x^2-19x+78), so (x^2-19x+78=22) gives (x^2-19x+56=0). In exams, bring all terms to one side after expansion.

Here (D=(-19)^2-4(1)(56)=137), so (x=\frac{19\pm\sqrt{137}}{2}). In exams, finding (D) correctly is important.