Methods of Solving Quadratic Equations

Here (ac=60) and (10+6=16), so (16x) is split as (10x+6x). In exams, check both sum (b) and product (ac).

(15x^2+16x+4=(3x+2)(5x+2)), so the roots are (-\frac{2}{3}) and (-\frac{2}{5}). In exams, write fractional roots in simplest form.

For equal roots, (D=0), so (4(k-1)^2-4(k^2-4)=0) gives (-2k+5=0). In exams, expand (D) carefully in parameter questions.

The sum of roots is (4), and (\frac{p+2}{3}=4), so (p=10). In exams, find the sum of roots using (-\frac{b}{a}).

First \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) is obtained, then \(\left(x-\frac{9}{5}\right)^2=\frac{36}{25}\). In exams, divide by (a) first when \(a\neq1\).

(5x^2-18x+9=(5x-3)(x-3)), so the roots are (\frac{3}{5}) and (3). In exams, verify the answer quickly by factorisation.

The direct answer is option A: \(m=49\). For a quadratic \(ax^2+bx+c=0\), real and equal roots occur exactly when the discriminant \(D=b^2-4ac\) is zero. Here \(a=1\), \(b=-14\), and \(c=m\). Thus \(D=(-14)^2-4(1)(m)=196-4m\). Set it equal to zero: \(196-4m=0\), so \(4m=196\), and \(m=49\). Option A is correct. Option B, \(m=14\), gives \(D=140\), so there are two distinct real roots. Option C, \(m=28\), gives \(D=84\), also two distinct real roots. Option D, \(m=196\), gives \(D=-588\), so the roots are not real. The key fact is: equal real roots mean the discriminant is exactly zero, not merely positive or negative.

The roots of the first equation are (2,\frac{4}{3}), and the roots of the second are (2,1). In exams, solve both equations separately for the common root.

The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.

The other root is (5), so (q=4\times5=20). In exams, (c) equals the product of roots when (a=1).

(3x^2+x-2=(3x-2)(x+1)), so (x=\frac{2}{3},-1) is correct. In exams, change signs carefully from factors.

Since (5=(\sqrt{5})^2) and the middle term is (-2\sqrt{5}x), it is ((x-\sqrt{5})^2). In exams, identify perfect squares even with irrational coefficients.

The equation is equivalent to ((x-m)(x-n)=0), so the roots are (m) and (n). In exams, the same rule applies to symbolic factors.

It is ((x-p)^2-q^2=0), so (x-p=\pm q) and (x=p\pm q). In exams, recognize the difference of squares.

Dividing the whole equation by (9) gives (x^2-(r+s)x+rs=0). In exams, removing the common factor first makes solving easier.

Since (x^4=(x^2)^2=y^2), the new equation is (y^2-10y+9=0). In exams, substitution can simplify a difficult form.

From (y^2-10y+9=0), (y=1,9), so (x^2=1,9) and (x=\pm1,\pm3). In exams, do not forget to return to (x).

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).