Methods of Solving Quadratic Equations

((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.

Split the middle term as \(-2x-5x\): \(2x^2-7x+5=2x^2-2x-5x+5=(2x-5)(x-1)\). Hence, the factorised equation is \((2x-5)(x-1)=0\). Option C is incorrect because its expansion gives the middle term \(-11x\), not \(-7x\). As an exam tip, expand the factors once to verify the middle term and constant term.

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.

(x^2-15x+56=(x-7)(x-8)), so the roots are (7) and (8). In exams, check both sum and product.

(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.