Methods of Solving Quadratic Equations

The sum of roots is (-\frac{b}{a}=-\frac{-13}{4}=\frac{13}{4}). In exams, keep the sign of (b) carefully.

The product of roots is (\frac{c}{a}=\frac{3}{4}). In exams, use (\frac{c}{a}) for the product.

\((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{13}{4}\right)^2-3=\frac{121}{16}\). In exams, use this identity for the square of difference.

For no real roots, (D<0), so (100-4n<0) and (n>25). In exams, connect (D<0) with no real roots.

For two distinct real roots, (D>0), so (100-4n>0) and (n<25). In exams, connect (D>0) with distinct real roots.

Expanding option A gives \((5x+3)(x-2)=5x^2-10x+3x-6=5x^2-7x-6\), so it is the correct factorised form. Option B produces a middle coefficient of \(+7\), while options C and D produce \(+13\) and \(-13\), respectively. In an exam, always expand the factors and compare the result with the original quadratic equation.

The quadratic factors as \(5x^2-7x-6=(5x+3)(x-2)\). Thus, \((5x+3)(x-2)=0\) gives \(x=-\frac{3}{5}\) or \(x=2\), so option A is correct. In option B, the signs of both roots are incorrect. Exam tip: when setting each factor equal to zero, remember to change the sign while solving for the root.

Completing the square means converting the quadratic expression into a perfect-square form while preserving equality. First move the constant term: x^2 + 8x = -5. Half the coefficient of x is 8/2 = 4, and its square is 16. Add 16 to both sides: x^2 + 8x + 16 = -5 + 16. The left side becomes (x+4)^2, so (x+4)^2 = 11. Hence option A is correct. Option B has the wrong sign inside the square, while options C and D either use the wrong coefficient or fail to add 16 correctly.

Completing the square in \(x^2+8x+5=0\) gives \(x^2+8x+16=11\), or \((x+4)^2=11\). Hence, \(x+4=\pm\sqrt{11}\), so the roots are \(x=-4\pm\sqrt{11}\). Option B has the wrong sign before 4, while option D incorrectly omits the square-root sign. Exam tip: the \(\pm\) symbol represents both roots, so write both values when required.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=6) and (\alpha\beta=-16), so the value is (-96). In exams, factor the expression first.

Let the roots be (-r) and (-2r), then (2r^2=25) and (p=3r=\frac{15\sqrt{2}}{2}). In exams, do not forget to rationalize the denominator.

Here (D=(-7)^2-4(1)(3)=37), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{37}). In exams, the difference of roots can be found directly from (D).

Here (D=(-8)^2-4(4)(9)=-80<0), so there are no real roots. In exams, (D<0) means no real roots.

(4x^2-8x+9=4(x-1)^2+5), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.

The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=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=14) and (\alpha\beta=45), so the value is (\frac{196-90}{45}=\frac{106}{45}). In exams, convert expressions into sum and product.

For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-14\), and \(c=49\), so \(D=(-14)^2-4(k)(49)=196-196k\). Thus, \(196-196k=0\), giving \(k=1\). Exam tip: identify the coefficient of \(x^2\) as \(a=k\); the equation is quadratic only when \(k\neq 0\).

(D=4(k+4)^2-4k^2=0) gives ((k+4)^2=k^2), so (8k+16=0) and (k=-2). In exams, expand squares carefully.

The governing principle is the zero-product rule: if the product of two real factors is zero, at least one factor must be zero. Applying it here gives x-4=0 or x-9=0. Solving the two simple linear equations separately gives x=4 or x=9. Therefore the solution set is {4,9}, and option A is correct. Option B changes both signs incorrectly; substituting x=-4 or x=-9 does not make the corresponding factors zero. Option C comes from unrelated addition or subtraction, and option D incorrectly treats the constants as if their product and sum were the roots. Factoring already provides the solutions directly.

((x-4)(x-9)=x^2-13x+36), so (x^2-13x+36=14) gives (x^2-13x+22=0). In exams, bring all terms to one side after expansion.