Methods of Solving Quadratic Equations

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.

Direct answer: option A, (x+5)²=19. Begin with x²+10x+6=0. Move the constant: x²+10x=−6. Half of the coefficient of x is 10/2=5, and its square is 25. Add 25 to both sides: x²+10x+25=−6+25=19. The left side is (x+5)², so (x+5)²=19. A is correct. B has x−5, which would produce x²−10x, not x²+10x. C incorrectly treats 10 as the number to add inside the square and leaves the constant handling wrong. D has the correct left expression but incorrectly keeps the right side as 6 instead of 19. Adding the same number to both sides preserves equality. Memory cue: half the x-coefficient, square it, add it on 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.

Let the roots be (-r) and (-2r), then (2r^2=36) gives (r=3\sqrt{2}), and (p=3r=9\sqrt{2}). In exams, keep signs of both roots carefully.

Here (D=(-9)^2-4(1)(5)=61), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{61}). In exams, the difference of roots can be found directly from (D).

Here (D=(-10)^2-4(5)(13)=-160<0), so there are no real roots. In exams, (D<0) means no real roots.

(5x^2-10x+13=5(x-1)^2+8), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.

The roots are (2,10), so new roots are (6,14), and the equation is ((x-6)(x-14)=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=18) and (\alpha\beta=77), so the value is (\frac{324-154}{77}=\frac{170}{77}). In exams, convert expressions into sum and product.

A quadratic equation \(ax^2+bx+c=0\) has equal roots when its discriminant \(b^2-4ac\) is zero. Here, \(a=k\), \(b=-18\), and \(c=81\). Thus, \((-18)^2-4(k)(81)=0\), giving \(324-324k=0\) and hence \(k=1\). Exam tip: identify the coefficient of \(x^2\) as \(a\) before applying the discriminant formula.

(D=4(k+5)^2-4k^2=0) gives ((k+5)^2=k^2), so (10k+25=0) and (k=-\frac{5}{2}). In exams, expand squares carefully.

Direct answer: option A, x=5 and x=11. Use the zero-product rule: if a product of two factors is zero, at least one factor must be zero. Thus (x−5)(x−11)=0 gives two cases. Case 1: x−5=0, so x=5. Case 2: x−11=0, so x=11. Therefore the solutions are 5 and 11. B gives negative values, but x−5=0 produces positive 5 and x−11=0 produces positive 11. C, 0 and 16, do not make either given factor zero. D, 1 and 55, are unrelated values and do not satisfy the factors. Checking confirms: at x=5 the first factor is zero; at x=11 the second factor is zero. Do not multiply first unless needed; separate factors directly. Memory cue: for (x−a)(x−b)=0, write x=a or x=b.

((x-5)(x-11)=x^2-16x+55), so (x^2-16x+55=18) gives (x^2-16x+37=0). In exams, bring all terms to one side after expansion.

Here (D=(-16)^2-4(1)(37)=108), so (x=\frac{16\pm6\sqrt{3}}{2}=8\pm3\sqrt{3}). In exams, simplify (D) correctly.