Methods of Solving Quadratic Equations

(x^2-11x=x(x-11)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.

(9+10=19) and (9\times10=90), so ((x-9)(x-10)) is correct. In exams, take both negative signs for a negative middle term.

((x-9)(x-10)=0), so (x=9) and (x=10). In exams, roots are obtained by taking opposite signs of factors.

((-9)+(-2)=-11) and ((-9)(-2)=18=ac), so (-9x-2x) is correct. In exams, split the middle term by checking (ac).

Expanding \((3x-2)(x-3)\) gives \(3x^2-9x-2x+6=3x^2-11x+6\), so option A is correct. Option C expands to \(3x^2-9x+6\), whose coefficient of \(x\) is incorrect. In an exam, expand the factors to check both the middle term and the constant term.

Factoring the quadratic gives \\(3x^2-11x+6=(3x-2)(x-3)\\). Therefore, \\(3x-2=0\\) gives \\(x=\frac{2}{3}\\), and \\(x-3=0\\) gives \\(x=3\\). Hence, option A is correct. Option D incorrectly uses \\(\frac{3}{2}\\) instead of \\(\frac{2}{3}\\). In an exam, set each factor equal to zero and solve for the roots separately.

Taking square roots gives \\(x-4=\\pm 5\\). Thus, \\(x-4=5\\) gives \\(x=9\\), while \\(x-4=-5\\) gives \\(x=-1\\). Option B results from handling the addition of 4 incorrectly. In exams, remember to include both \\(\\pm\\) cases when taking a square root.

The coefficient of \(x\) is 24, so its half is 12. To complete the square, we add the square of this half: \(\left(\frac{24}{2}\right)^2=12^2=144\). Therefore, 144 is added to both sides. Remember that 12 is the half of the coefficient, while 144 is the number added.

Using (a + b)² = a² + 2ab + b², we get (x + 12)² = x² + 2(x)(12) + 12² = x² + 24x + 144. Therefore, option A is correct. The minus sign in (x − 12)² would instead produce a middle term of −24x.

(11+(-5)=6) and (11\times(-5)=-55), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.

((x+11)(x-5)=0), so (x=-11) and (x=5). In exams, the sign changes while finding roots from factors.

Dividing both sides of the equation by 12 gives \(\frac{12x^2}{12}=\frac{108}{12}\), so \(x^2=9\). Solving further gives \(x=\pm3\). In an exam, remember to perform the same operation on both sides of an equation.

Dividing both sides by 12 gives \(x^2=9\). Therefore, \(x=\pm\sqrt{9}=\pm3\), so the two solutions are \(x=3\) and \(x=-3\). Writing only \(x=3\) is incomplete because the square of \(-3\) is also 9. In exams, remember to include both the positive and negative square roots.

In the quadratic formula, the part inside the square root is (b^2-4ac). In exams, it is also called the discriminant (D).

\((x+5)^2=x^2+2(5)x+5^2=x^2+10x+25\). Comparing this with \(x^2+10x+k\) gives \(k=25\). The distractor 5 is the constant inside the bracket, whereas \(k\) is its square. Exam tip: in \((x+a)^2=x^2+2ax+a^2\), the constant term is \(a^2\).

In completing square method, the quadratic part is made into the form ((x+p)^2). In exams, this method is useful when simple factors are not found quickly.

(25x^2-1=(5x)^2-1^2), so ((5x-1)(5x+1)=0) is correct. In exams, quickly identify the difference of squares.

The equation is most efficiently solved by factorisation. We look for two binomials whose product has first term 5x², constant term 1, and middle term 6x. The factorisation is 5x² + 6x + 1 = (5x + 1)(x + 1), because multiplication gives 5x² + 5x + x + 1 = 5x² + 6x + 1. By the zero-product property, either 5x + 1 = 0 or x + 1 = 0. These equations give x = −1/5 and x = −1, respectively. Therefore option A is correct. Option B changes both signs, option C treats coefficients as roots rather than solving the factors, and option D gives positive values that do not satisfy the original equation. Direct substitution confirms both negative roots.

The governing algebraic identity is (p−q)²=p²−2pq+q². In the given quadratic, 16x² is (4x)² and the constant term 9 is 3². Taking p=4x and q=3 gives (4x−3)²=(4x)²−2(4x)(3)+3²=16x²−24x+9. Hence the equation is exactly (4x−3)²=0, so option A is correct. Expanding option B would produce a positive middle term, +24x, rather than −24x. Option C has leading term 256x², and option D has leading term x², so neither can represent the original expression. The identity also shows that the repeated root would be x=3/4, although finding the root is not needed to identify the form.

A quadratic equation is commonly written in standard form as ax² + bx + c = 0, with a ≠ 0. Starting from 5x² + 9x = 2, subtract 2 from both sides to preserve equality: 5x² + 9x − 2 = 2 − 2 = 0. Thus the standard-form equation is 5x² + 9x − 2 = 0, which is option A. Option B results from changing the sign incorrectly while moving 2. Option C changes the sign of the 9x term without justification, and option D interchanges the coefficients of x² and x. No solving of the quadratic is required; the task only asks for rearrangement into standard form.

The governing concept is the identity (p − q)² = p² − 2pq + q². Choose p = 6x and q = 5. Then (6x − 5)² = (6x)² − 2(6x)(5) + 5² = 36x² − 60x + 25. Thus the given quadratic equation is exactly equivalent to (6x − 5)² = 0, so option A is correct. The plus form in option B would produce a positive middle term, +60x. In option C, the leading term would be (36x)² = 1296x², not 36x². In option D, the leading term would be x². Matching the leading, middle, and constant coefficients confirms that only option A represents the given perfect square.

To solve an equation containing a square, take both square roots because a real number can have either a positive or a negative square root. From (x−7)²=11, we obtain x−7=±√11. Adding 7 to both sides gives x=7±√11. Thus the two solutions are x=7+√11 and x=7−√11, making option A correct. Option B incorrectly changes the sign of 7 when it is transposed. Option C uses 11 instead of its square root, and option D places the square root on the wrong number and also omits the correct structure. The ± symbol is essential because both values square to 11.

The governing concept is the zero-product rule: if the product of two real expressions is zero, at least one factor must be zero. Therefore, from (x−3)(x−7)=0, set the factors separately equal to zero: x−3=0 gives x=3, and x−7=0 gives x=7. Thus the solution set is {3,7}, so option A is correct. Option B incorrectly changes both signs, while options C and D do not make either original factor zero in the required alternatives. Substitution confirms both answers: for x=3 the first factor is zero, and for x=7 the second factor is zero.

The governing principle is the zero-product rule: if the product of two factors is zero, at least one factor must be zero. Applying it to (x-4)(x-9)=0 gives two cases: x-4=0 or x-9=0. Solving them separately gives x=4 and x=9, so the solution set is {4,9}. Therefore option A is correct. Option B changes the signs incorrectly; substituting -4 or -9 does not make the corresponding factors zero. Option C has no valid connection with the factorized equation, and option D incorrectly treats the constants as though their sum and product were the roots. Since the expression is already factorized, the zero-product rule gives the solutions directly and no quadratic formula is needed.