Methods of Solving Quadratic Equations

The product of roots is (\frac{p}{10}), so (\frac{p}{10}=\frac{2}{5}) gives (p=4). In exams, use the product formula (\frac{c}{a}).

(\alpha+\beta=27) and (\alpha\beta=180), so (\alpha^2+\beta^2=27^2-2(180)=369). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}). In exams, write the answer in simplest form.

The sum of roots is (-\frac{b}{a}=-\frac{-31}{8}=\frac{31}{8}). In exams, keep the sign of (b) carefully.

The product of roots is (\frac{c}{a}=\frac{15}{8}). In exams, use (\frac{c}{a}) for the product.

\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{31}{8}\right)^2-\frac{15}{2}=\frac{481}{64}\). In exams, convert fractions to a common denominator.

For no real roots, (D<0), so (256-4n<0) and (n>64). In exams, connect (D<0) with no real roots.

For two distinct real roots, (D>0), so (256-4n>0) and (n<64). In exams, connect (D>0) with distinct real roots.

((8x+5)(x-3)=8x^2-19x-15), so it is not for the given equation. In exams, verify each option by expansion.

((x+5)(8x-3)) does not expand to the given equation, so the options must be checked carefully. The correct factorisation is not present among careless options.

((4x+3)(2x-5)=8x^2-20x+6x-15=8x^2-14x-15), so it is correct. In exams, verify factorisation by expanding.

((4x+3)(2x-5)=0), so (x=-\frac{3}{4}) and (\frac{5}{2}). In exams, change signs while writing roots.

The direct answer is A,
(x+7)^2=39
. Begin with
x^2+14x+10=0
. Move the constant term to the other side:
x^2+14x=-10
. Half of the coefficient of
x
is
7
, and its square is
49
. Add
49
to both sides:
x^2+14x+49=-10+49
. The left side becomes
(x+7)^2
, so the result is
(x+7)^2=39
. Option A is therefore correct. Option B has
x-7
; expanding it gives
x^2-14x+49
, so its middle sign is wrong. Option C uses
(x+14)^2
, whose expansion has
28x
, not
14x
, and it also keeps the wrong right side. Option D has the correct square form but the right side is incorrectly written as
10
instead of
39
. Completing the square always requires adding the same number to both sides. Memory cue: half the linear coefficient, square it, add it to both sides.

Rewrite the quadratic by completing the square: \(x^2+14x+10=(x+7)^2-39\). Thus, \((x+7)^2=39\), so \(x+7=\pm\sqrt{39}\) and \(x=-7\pm\sqrt{39}\). Option B has the wrong sign for the constant term in the root expression, while option D omits the square root. In an exam, remember that \(\pm\) represents both roots.

(\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.