Methods of Solving Quadratic Equations

The product of roots is (\frac{p}{9}), so (\frac{p}{9}=\frac{1}{3}) gives (p=3). In exams, use the product formula (\frac{c}{a}).

(\alpha+\beta=23) and (\alpha\beta=126), so (\alpha^2+\beta^2=23^2-2(126)=277). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}). In exams, first write sum and product in reciprocal questions.

The sum of roots is (-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}). In exams, keep the sign of (b) carefully.

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.

Direct answer: option A, (x+6)²=28. Start with x²+12x+8=0 and move 8 to the other side: x²+12x=−8. Half of 12 is 6, and 6²=36. Add 36 to both sides: x²+12x+36=−8+36=28. The left side is (x+6)², so the required form is (x+6)²=28. B has the wrong sign: (x−6)² gives −12x, not +12x. C uses x+12 instead of x+6 and does not complete the square correctly. D has the correct square on the left but fails to add 36 on the right, so 8 is not correct. The same quantity must be added to both sides. Memory cue: use half of the x coefficient, not the full coefficient.

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.

Direct answer: option A, x=6 and x=13. Apply the zero-product rule to (x−6)(x−13)=0. Either x−6=0 or x−13=0. From the first equation, adding 6 gives x=6. From the second, adding 13 gives x=13. Both values work: x=6 makes the first factor zero, and x=13 makes the second factor zero. B has incorrect negative signs; the equations contain x minus positive numbers, so their zeros are positive 6 and 13. C, 0 and 19, are not obtained by setting either factor to zero. D, 1 and 78, also do not make either factor zero. A product can be zero when one factor is zero; both factors need not be zero at the same time. Memory cue: set each bracket separately equal to zero, then solve.

((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.

First (x^2+6x+\frac{5}{3}=0) is obtained, then adding (9) gives ((x+3)^2=\frac{22}{3}). In exams, divide by (a) first when (a\neq1).