Methods of Solving Quadratic Equations

From (y^2-17y+16=0), (y=1,16), so (x^2=1,16) and (x=\pm1,\pm4). In exams, do not forget to return to (x).

Multiplying both sides by (5x) gives (5+5x^2=26x), that is (5x^2-26x+5=0). In exams, remember the condition (x\neq0).

(5x^2-26x+5=(5x-1)(x-5)), so (x=\frac{1}{5}) and (5). In exams, check whether obtained roots are valid in the original equation.

Cross multiplication gives ((x+4)^2=25x), so (x^2+8x+16-25x=0), and (x^2-17x+16=0). In exams, cross multiply carefully.

(x^2-17x+16=(x-1)(x-16)), so (x=1) and (x=16). In exams, check solutions against excluded denominator values.

(D=(-10)^2-4(1)(7)=72), so (x=\frac{10\pm6\sqrt{2}}{2}=5\pm3\sqrt{2}). In exams, simplify the square root.

The sum of roots is (-\frac{p}{6}), so (-\frac{p}{6}=-8) gives (p=48). In exams, remember the sum formula (-\frac{b}{a}).

The product of roots is (\frac{p}{7}), so (\frac{p}{7}=\frac{2}{7}) gives (p=2). In exams, use the product formula (\frac{c}{a}).

(\alpha+\beta=19) and (\alpha\beta=88), so (\alpha^2+\beta^2=19^2-2(88)=185). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{16}{63}). In exams, first write sum and product in reciprocal questions.

The sum of roots is (-\frac{b}{a}=-\frac{-17}{5}=\frac{17}{5}). In exams, keep the sign of (b) carefully.

The product of roots is (\frac{c}{a}=\frac{6}{5}). In exams, use (\frac{c}{a}) for the product.

\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{17}{5}\right)^2-\frac{24}{5}=\frac{169}{25}\). In exams, convert fractions to a common denominator.

For no real roots, (D<0), so (144-4n<0) and (n>36). In exams, connect (D<0) with no real roots.

For two distinct real roots, (D>0), so (144-4n>0) and (n<36). In exams, connect (D>0) with distinct real roots.

Expanding option A gives \((3x+2)(2x-5)=6x^2-15x+4x-10=6x^2-11x-10\), so it is the correct factorised form. In option B, the middle term becomes \(+11x\), making it incorrect. In an exam, verify factorisation by expanding the factors and comparing the result with the original polynomial.

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

Adding (25) to (x^2+10x=-6) gives ((x+5)^2=19). In exams, add the same number to both sides.

Here, \(a=1, b=10, c=6\). Using the quadratic formula, \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-10\pm\sqrt{100-24}}{2}=\frac{-10\pm\sqrt{76}}{2}=-5\pm\sqrt{19}\). Therefore, option A is correct. Option B has the wrong sign for \(-b\), while option D omits the square root of the simplified discriminant. Exam tip: verify that the sum of the roots is \(-10\) and their product is \(6\).

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=8) and (\alpha\beta=-20), so the value is (-160). In exams, factor the expression first.