Methods of Solving Quadratic Equations

(x^2-5x+6=(x-2)(x-3)), so the roots are (2) and (3). In exams, first find two numbers whose product is (6) and sum is (-5).

Because (3+4=7) and (3\times4=12), the correct factors are ((x+3)(x+4)). In exams, pay close attention to signs.

(x^2-9=x^2-3^2=(x-3)(x+3)), so it is solved quickly by difference of squares. In exams, recognizing (a^2-b^2) is useful.

From (x^2=16), (x=\pm\sqrt{16}=\pm4). In exams, do not forget (\pm) while taking square root.

The zero-product rule states that if the product of two real factors is zero, then at least one factor must be zero. In x(x−4)=0, set the first factor equal to zero: x=0. Then set the second factor equal to zero: x−4=0, which gives x=4. Hence the solution set is {0,4}, so option A is correct. Substitution confirms both results: 0·(0−4)=0 and 4·(4−4)=0. Option B comes from changing the sign incorrectly and solving x+4=0, which is not the given factor. Option C and option D do not make either factor zero in the required way. Factoring has already been done, so applying the quadratic formula is unnecessary.

Half of the coefficient (6) is (3), and \(3^2=9\). In exams, use \(\left(\frac{b}{2}\right)^2\).

The quadratic formula is (x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}). In exams, identifying (a), (b), and (c) correctly is most important.

Dividing both sides by (2) gives (x^2-4=0). In exams, simplifying the equation first saves time.

Since (-4+2=-2) and (-4\times2=-8), ((x-4)(x+2)) is correct. In exams, match both product and sum.

(x^2+4x+4) is a perfect square and equals ((x+2)^2). In exams, recognize the pattern (a^2+2ab+b^2).

(x^2-6x+9=(x-3)^2), so the repeated root is (3). In exams, a perfect square gives equal roots.

The quadratic formula is applied to an equation in standard form ax²+bx+c=0. Compare 3x²+5x−2=0 term by term with that form. The coefficient of x² is a, so a=3; the coefficient of x is b, so b=5; and the constant term is c, so c=−2. Therefore option A is correct. The negative sign belongs to the constant term and must not be lost. Option B swaps the coefficients of x² and x. Option C changes both signs of the linear and constant terms, although the original equation has +5x and −2. Option D assigns the constant to a and the leading coefficient to c, reversing the standard roles. Correct identification of these coefficients is essential before substitution into the formula x=[−b±√(b²−4ac)]/(2a).

The terms x² and 5x have x as a common factor. Factoring it out gives x² + 5x = x(x + 5), so the equation becomes x(x + 5) = 0. Option B has the wrong sign; the correct rearrangement would be x² = -5x. Exam tip: factor out the common term first, then apply the zero-product property to set each factor equal to zero.

It is ((x+5)^2=0), so recognizing the perfect square is fastest. In exams, look for patterns to save time.

(x^2-7x+10=(x-2)(x-5)), so the roots are (2) and (5). In exams, check (2+5=7) and (2\times5=10).

(5+(-3)=2) and (5\times(-3)=-15), so this pair is correct. In exams, split the middle term using such a pair.

For a quadratic equation written in the standard form ax² + bx + c = 0, the letters a, b and c represent the coefficients of x², x and the constant term respectively. Comparing 2x² + 7x + 3 = 0 with this form gives a = 2, b = 7 and c = 3. Therefore, ac means the product of the first and last coefficients: ac = 2 × 3 = 6. This product is useful because the middle term 7x can then be split into two terms whose coefficients have product 6 and sum 7, namely 6x and x. Hence option A is correct. The values 7 and 3 are individual coefficients, while 10 is obtained by adding a and c, not multiplying them.

Since (6+1=7) and (6\times1=6), split (7x) as (6x+x). In exams, keep the sum (b) and product (ac).

The equation gives \(x^2=25=5^2\). Therefore, \(x=5\) or \(x=-5\), that is, \(x=\pm5\). Options B and C incorrectly treat 25 as a root, while option D uses \(\pm25\) instead of \(\pm\sqrt{25}\). In an exam, remember that \(x^2=a^2\) has the two solutions \(x=\pm a\).

The governing idea is to take the greatest common factor from every term. In 3x² − 12x = 0, both terms contain 3x. Dividing each term by 3x leaves x and −4, so 3x² − 12x = 3x(x − 4). Retaining the equation gives 3x(x − 4) = 0, which is option A. This form is especially useful because the zero-product property can later be applied: either 3x = 0 or x − 4 = 0. Option B incorrectly removes x from the common factor, option C changes the sign of the second term, and option D gives the wrong sign inside the bracket. Factoring must preserve the original expression exactly.