Methods of Solving Quadratic Equations

\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{9}{2}\right)^2-8=\frac{65}{4}\). In exams, use this identity for square of difference.

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

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

Expanding (2x+1)(x−2) gives 2x² − 4x + x − 2 = 2x² − 3x − 2, so option A is correct. The closest distractor is option B, whose expansion is 2x² + 3x − 2; the sign of the coefficient of x is wrong. In an exam, expand the factors and compare the result with the original equation.

Begin by moving the constant term to the right: x² + 4x = −1. To complete the square, add (4/2)² = 2² = 4 to both sides. The left side becomes x² + 4x + 4 = (x+2)², while the right side becomes −1+4=3. Therefore the equivalent equation is (x+2)²=3, so option A is correct. Option B has the wrong sign inside the square and would create −4x. Option C adds an incorrect amount and has the wrong coefficient structure, while option D fails to account for the added 4 on the right. Adding the same quantity to both sides preserves equivalence.

Rewrite the equation by completing the square: \(x^2+4x+1=0\Rightarrow x^2+4x+4=3\Rightarrow (x+2)^2=3\). Thus, \(x+2=\pm\sqrt{3}\), giving \(x=-2\pm\sqrt{3}\). Option B has the wrong sign for the constant term in the solution. In an exam, remember that \(\pm\) represents both roots.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=2) and (\alpha\beta=-8), so the value is (-16). In exams, factor the expression first.

Let the roots be (-r) and (-2r), then (2r^2=9) and the sum is (-3r), so (p=3r=\frac{9}{\sqrt{2}}). In exams, assume the roots and form equations carefully.

(\frac{9}{\sqrt{2}}) simplifies to (\frac{9\sqrt{2}}{2}). In exams, do not forget to rationalize the denominator.

The square of the difference is (D/a^2=5), so the difference is (\sqrt{5}). In exams, the difference of roots is (\frac{\sqrt{D}}{|a|}).

Here (D=(-4)^2-4(2)(5)=-24<0), so there are no real roots. In exams, (D<0) means no real roots.

(2x^2-4x+5=2(x-1)^2+3), so it cannot be zero for real (x). In exams, completed square form also shows no real roots.

The roots are (1,5), so new roots are (2,6), and the equation is ((x-2)(x-6)=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=8) and (\alpha\beta=15), so the value is (\frac{64-30}{15}=\frac{34}{15}). In exams, convert expressions into sum and product.

For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-6\), and \(c=9\), so \(D=(-6)^2-4(k)(9)=36-36k\). Thus, \(36-36k=0\), giving \(k=1\). Exam tip: identify \(a\) as \(k\), since it is the coefficient of \(x^2\).

(D=4(k+2)^2-4k^2=0) gives ((k+2)^2=k^2), so (4k+4=0) and (k=-1). In exams, expand squares carefully.

Putting (m=0) gives ((x-2)(x-5)=0), so (x=2) or (x=5). In exams, apply zero product rule directly.

((x-2)(x-5)=x^2-7x+10), so (x^2-7x+10=6) gives (x^2-7x+4=0). In exams, bring all terms to one side after expansion.

Here (D=(-7)^2-4(1)(4)=33), so (x=\frac{7\pm\sqrt{33}}{2}). In exams, finding (D) correctly is important.

First (x^2+4x+\frac{1}{2}=0) is obtained, then adding (4) gives ((x+2)^2=\frac{7}{2}). In exams, divide by (a) first when (a\neq1).