Methods of Solving Quadratic Equations

((3x-4)(3x+4)=0), so (x=\frac{4}{3}) or (x=-\frac{4}{3}). In exams, solve linear factors carefully.

The direct answer is option A: a²−b²=(a−b)(a+b). Rewrite 100 as 10². Then x²−100=x²−10², which has the form a²−b² with a=x and b=10. Therefore the identity factors it as (x−10)(x+10), and the equation becomes (x−10)(x+10)=0. Option A applies exactly because the two square terms are separated by subtraction. Option B is the identity for the square of a sum and would require a middle term +2ab. Option C is for the square of a difference and would contain a middle term −2ab. Option D is false in general: a²+b² is not equal to (a+b)², because the latter also contains 2ab. For example, x²−100=0 gives x=±10 after factoring. Exam cue: two perfect squares with a minus sign indicate difference of squares.

(x^2+8x+12=(x+2)(x+6)), so the roots are (-2) and (-6). In exams, a positive middle term can give negative roots.

(x^2-5x=x(x-5)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.

(5+8=13) and (5\times8=40), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.

((x-5)(x-8)=0), so (x=5) and (x=8). In exams, roots are obtained by taking opposite signs of factors.

((-2)+(-5)=-7) and ((-2)(-5)=10=ac), so (-2x-5x) is correct. In exams, checking (ac) is important while splitting the middle term.

The correct answer is A. Expanding \((2x-5)(x-1)\) gives \(2x^2-2x-5x+5=2x^2-7x+5\), which matches the quadratic expression in the equation. Therefore, its factorised form is \((2x-5)(x-1)=0\).

The factorisation is \(2x^2-7x+5=(2x-5)(x-1)\). Therefore, \((2x-5)(x-1)=0\) gives \(x=\frac{5}{2}\) or \(x=1\). The values in option D do not satisfy the original equation. As an exam tip, substitute the obtained roots back into the equation to verify them.

Taking the square root gives \\(x+3=\\pm4\\). Thus, \\(x+3=4\\) gives \\(x=1\\), while \\(x+3=-4\\) gives \\(x=-7\\). Therefore, option A is correct. The values in option B do not satisfy the original equation. In an exam, remember to consider both the positive and negative square-root cases.

The coefficient of \(x\) is 16. Its half is 8, and the square of 8 is \(8^2=64\). Thus, adding 64 to both sides gives \(x^2+16x+64=(x+8)^2\). The number 8 is only half the coefficient, not the number to be added. Exam tip: For \(x^2+bx\), add \(\left(\frac{b}{2}\right)^2\) to complete the square.

\\((x+8)^2=x^2+2\cdot8x+8^2=x^2+16x+64\\), so option A is correct. Option B would produce the middle term \\-16x\\), not the given \\(+16x\\). In an exam, identify a perfect square by taking the square root of the constant term and checking whether twice that number gives the middle-term coefficient.

(7+(-2)=5) and (7\times(-2)=-14), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.

((x+7)(x-2)=0), so (x=-7) and (x=2). In exams, the sign changes while finding roots from factors.

Dividing both sides of \(8x^2=72\) by 8 gives \(x^2=9\). Taking square roots then gives \(x=\pm3\). Option B is incorrect because 72 has not been divided by 8. In an exam, first remove the coefficient of \(x^2\) by performing the same operation on both sides.

Dividing both sides by 8 gives \(x^2=9\). Therefore, \(x=\pm\sqrt{9}=\pm3\), so both \(x=3\) and \(x=-3\) are solutions. Writing only \(x=3\) or only \(x=-3\) is incomplete. Exam tip: remember to include both positive and negative values when taking the square root of a positive number.

In (x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}), the denominator is (2a). In exams, forgetting (2a) is a common mistake.

Expanding the perfect square gives \((x-3)^2=x^2-6x+9\). Comparing this with \(x^2-6x+k\), the constant term is \(k=9\). Therefore, option A is correct. Exam tip: in \((x-a)^2=x^2-2ax+a^2\), the constant term is \(a^2\).

In the quadratic formula, (b^2-4ac) is used as the discriminant. In exams, identify (a), (b), and (c) before using the formula method.

(4x^2+4x+1=(2x+1)^2), so it is a perfect square equation. In exams, recognize ((a+b)^2) to solve quickly.

The direct answer is option A: x=7 and x=8. Factorisation means writing the quadratic as a product of simpler factors. We need two numbers whose product is 56 and whose sum is 15, because the middle term is −15x. The numbers are 7 and 8. Thus x²−15x+56=(x−7)(x−8). Setting the product equal to zero gives (x−7)(x−8)=0, so either x−7=0 or x−8=0. Hence x=7 or x=8. Option A is correct. Option B, −7 and −8, would arise from factors (x+7)(x+8), whose middle term is +15x, not −15x. Option C, 4 and 14, has product 56 but sum 18, so it cannot produce the required middle term. Option D, 1 and 56, has product 56 but sum 57, so it also fails. Check both conditions—product 56 and sum 15—to avoid guessing.

(6+7=13) and (6\times7=42), so the correct factors are ((x+6)(x+7)). In exams, match the signs carefully.

(x^2-49=x^2-7^2), so the difference of squares method is fastest. In exams, recognizing (a^2-b^2) is useful.

Taking square roots on both sides of \(x^2=144\) gives \(x=\pm\sqrt{144}=\pm12\). Thus, the two solutions are \(x=12\) and \(x=-12\). Options B and C give only one solution each, while option D uses an incorrect value for \(\sqrt{144}\). Exam tip: when solving \(x^2=a\) for positive \(a\), remember both roots, \(x=\pm\sqrt{a}\).

((x-9)=0) or ((x+2)=0), so (x=9) or (x=-2). In exams, set each factor equal to zero separately.