Methods of Solving Quadratic Equations

Multiplying both sides by (3x) gives (3+3x^2=10x), that is (3x^2-10x+3=0). In exams, remember the condition (x\neq0).

(3x^2-10x+3=(3x-1)(x-3)), so (x=\frac{1}{3}) and (3). In exams, check whether obtained roots are valid in the original equation.

Cross multiplication gives ((x+2)^2=9x), so (x^2+4x+4-9x=0), and (x^2-5x+4=0). In exams, cross multiply carefully.

(x^2-5x+4=(x-1)(x-4)), so (x=1) and (x=4). In exams, check solutions against excluded denominator values.

(D=(-6)^2-4(1)(2)=28), so (x=\frac{6\pm2\sqrt{7}}{2}=3\pm\sqrt{7}). In exams, simplify the square root.

The sum of roots is (-\frac{p}{4}), so (-\frac{p}{4}=-6) gives (p=24). In exams, remember the sum formula (-\frac{b}{a}).

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

(\alpha+\beta=13) and (\alpha\beta=40), so (\alpha^2+\beta^2=13^2-2(40)=89). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}). In exams, first write sum and product in reciprocal questions.

The sum of roots is (-\frac{b}{a}=-\frac{-11}{3}=\frac{11}{3}). In exams, keep the sign of (b) carefully.

The product of roots is (\frac{c}{a}=\frac{6}{3}=2). In exams, use (\frac{c}{a}) for the product.

\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{11}{3}\right)^2-8=\frac{49}{9}\). In exams, use this identity for the square of difference.

For no real roots, (D<0), so (64-4n<0) and (n>16). In exams, connect (D<0) with no real roots.

For two distinct real roots, (D>0), so (64-4n>0) and (n<16). In exams, connect (D>0) with distinct real roots.

Expanding option A gives (3x+1)(x−2) = 3x² − 6x + x − 2 = 3x² − 5x − 2, so it is the correct factorised form. Option B gives a middle term of +5x, option C gives x, and option D gives −7x. In an exam, expand the factors and compare the result with the original quadratic equation.

The factorisation is \(3x^2-5x-2=(3x+1)(x-2)\). By the zero-product rule, \(3x+1=0\) or \(x-2=0\), giving the roots \(x=-\frac{1}{3}\) and \(x=2\). Option B has the signs of both roots reversed. As an exam check, the sum of the roots should be \(\frac{5}{3}\) and their product should be \(-\frac{2}{3}\).

The governing concept is completing the square while preserving equality. Start with x² + 6x + 2 = 0 and move the constant term to the other side: x² + 6x = −2. Half the coefficient of x is 6/2 = 3, and its square is 9. Add 9 to both sides, not just one side: x² + 6x + 9 = −2 + 9. The left side becomes (x + 3)² and the right side becomes 7, so (x + 3)² = 7. Therefore option A is correct. Option B has the wrong sign inside the square, option C uses 6 instead of half of 6, and option D adds 9 incorrectly or fails to account for the moved constant.

Rewrite the equation by completing the square: \(x^2+6x+2=0\Rightarrow (x+3)^2=7\). Thus, \(x+3=\pm\sqrt{7}\), giving \(x=-3\pm\sqrt{7}\). Option B has the wrong sign before 3, so it does not satisfy the equation. In an exam, remember that \(\pm\) represents both roots.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=4) and (\alpha\beta=-12), so the value is (-48). In exams, factor the expression first.

Let the roots be (-r) and (-2r), then (2r^2=16) gives (r=2\sqrt{2}), and (p=3r=6\sqrt{2}). In exams, keep signs of both roots carefully.