Methods of Solving Quadratic Equations

In \(x^2+8x+16\), the coefficient of \(x\) is 8. Taking half of 8 gives 4, and adding its square, 16, gives \(x^2+8x+16=(x+4)^2\). Therefore, A is correct. In \((x+8)^2\), the middle term would be \(16x\), so B is incorrect. Exam tip: To complete the square in \(x^2+bx\), add \(\left(\frac{b}{2}\right)^2\).

Using the square root method, \(x^2=49\) gives \(x=\pm\sqrt{49}\). Hence, \(x=7\) or \(x=-7\), so \(x=\pm7\). Writing only the positive root (option B) is a common mistake because a positive number has two square roots. Exam tip: For \(x^2=a\), always write \(x=\pm\sqrt{a}\).

Here, \(4x^2-1=(2x)^2-1^2\). Using the difference-of-squares identity \(a^2-b^2=(a-b)(a+b)\), it becomes \((2x-1)(2x+1)\), so option A is correct. Option C expands to \(4x^2-4x+1\), not \(4x^2-1\). In an exam, first check whether the expression matches the difference-of-squares pattern.

((2x-1)(2x+1)=0), so (x=\frac{1}{2}) or (x=-\frac{1}{2}). In exams, solve each linear factor carefully.

The expression x² − 1 is a difference of two perfect squares because 1 = 1². The identity for this pattern is a² − b² = (a − b)(a + b). Taking a = x and b = 1 gives x² − 1² = (x − 1)(x + 1). Hence the equation can be rewritten as (x − 1)(x + 1) = 0, leading to x = 1 or x = −1 by the zero-product property. Therefore option A is correct. Option B is the identity for the square of a sum and would require a middle term 2ab. Option C is not a valid identity in general, and option D is not the zero-product rule; from ab = 0, the correct conclusion is a = 0 or b = 0.

(x^2+6x+8=(x+2)(x+4)), so (x=-2) and (x=-4). In exams, a positive middle term can give negative roots.

(x^2-3x=x(x-3)), so zero product rule gives (x=0). In exams, do not lose (x=0) by dividing by (x).

First factor the equation without dividing by a variable: x² − 3x = x(x − 3) = 0. By the zero-product property, either x = 0 or x − 3 = 0, which gives x = 3. If we divide the original equation by x, we implicitly assume x is nonzero, so the solution x = 0 is excluded before the solving process is complete. The reduced equation x − 3 = 0 therefore retains only x = 3 and misses x = 0. Thus option A is correct. The safe procedure is to factor first and then set each factor equal to zero. Options B is a valid remaining root, while −3 and 1 do not satisfy the original equation.

(3+5=8) and (3\times5=15), so with signs we get ((x-3)(x-5)). In exams, for a negative middle term, look for both negative factors.

((-2)+(-3)=-5) and ((-2)(-3)=6=ac), so the correct split is (-2x-3x). In exams, the product of the two numbers must be (ac).

Multiplying \((2x-3)(x-1)\) gives \(2x^2-2x-3x+3=2x^2-5x+3\), so the correct factorised form is \((2x-3)(x-1)=0\). In option B, the coefficient of the middle term becomes \(1\) instead of \(-5\), while options C and D do not produce the required leading and constant terms. In an exam, multiply the factors back to verify the middle and constant terms.

In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\text{expression})^2=k).

Taking the square root gives \\(x-2=\\pm3\\). Thus, \\(x-2=3\\) gives \\(x=5\\), while \\(x-2=-3\\) gives \\(x=-1\\). Option D is incorrect because the second value should be \\(x=-1\\), not \\(x=1\\). In an exam, remember to consider both cases represented by \\(\\pm\\).

The coefficient of \(x\) is \(-10\), whose half is \(-5\). Squaring it gives \((-5)^2=25\), so 25 must be added: \(x^2-10x+25=(x-5)^2\). Exam tip: in the completing-square method, add the square of half the coefficient of \(x\).

Using the perfect-square identity \((x-a)^2=x^2-2ax+a^2\) with \(a=5\), we get \((x-5)^2=x^2-10x+25\). Hence, option A is correct. Option B would produce a middle term of \(+10x\), while option D represents a difference of squares. In an exam, identify \(25=5^2\) and verify the middle term as \(-2\times5x=-10x\).

(3+(-2)=1) and (3\times(-2)=-6), so ((x+3)(x-2)) is correct. In exams, match signs with the product.

((x+3)(x-2)=0), so (x=-3) and (x=2). In exams, the sign changes while finding roots from factors.

(x^2=-9) gives no solution in real numbers. In exams, remember that (x^2) cannot be negative for real (x).

Dividing both sides by (5) gives (x^2=4). In exams, remove the coefficient first and then take square root.

Dividing both sides of \(5x^2=20\) by 5 gives \(x^2=4\). Taking square roots, \(x=\pm\sqrt{4}=\pm2\), so the solutions are \(x=2\) and \(x=-2\). Options B and C give only one of the two roots, while option D results from an incorrect square root. Exam tip: remember to include both signs when taking the square root of a positive number.

The direct answer is option A: (x − 6)² = 0. To form a square, use (x − a)² = x² − 2ax + a². Here the constant term is 36, so a = 6 because 6² = 36. The middle term then becomes −2·6·x = −12x, exactly as required. Therefore x² − 12x + 36 = (x − 6)², and the equation is (x − 6)² = 0. Option A has both the correct sign and the correct number. Option B, (x + 6)², expands to x² + 12x + 36, so its middle sign is wrong. Option C, (x − 12)², has constant term 144 and middle term −24x. Option D, (x + 12)², has constant term 144 and a positive middle term. The reliable check is to square the proposed binomial before choosing. Memory cue: a negative middle term means the binomial has a minus sign.

((x-6)^2=0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.

The common factor in (x^2-16x) is (x), so we write (x(x-16)=0). In exams, do not divide by the variable and lose (x=0).

The governing concept is factorisation together with the zero-product property. We seek two numbers whose product is 20 and whose sum is 9; these numbers are 4 and 5. Therefore x² − 9x + 20 can be written as x² − 4x − 5x + 20 = (x − 4)(x − 5). The equation becomes (x − 4)(x − 5) = 0. By the zero-product property, either x − 4 = 0 or x − 5 = 0, giving x = 4 or x = 5. Option B would produce negative roots and a positive sum in the factorised expression, not the required middle coefficient. Options C and D have products or sums that do not match 20 and 9. Hence option A is correct.

(5+6=11) and (5\times6=30), so the correct factors are ((x+5)(x+6)). In exams, positive (c) and positive (b) give both positive signs.