Methods of Solving Quadratic Equations

(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.

(x^2-12x+36=(x-6)^2), so it is a perfect square. In exams, check (36=6^2) and (-12x=-2\cdot6x).

((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 idea is the zero-product property: if a product of two factors is zero, at least one factor must be zero. We need two numbers whose product is 20 and whose sum is 9. Those numbers are 4 and 5. Therefore, x² − 9x + 20 can be rewritten as (x − 4)(x − 5) = 0. Setting each factor equal to zero gives x − 4 = 0, so x = 4, and x − 5 = 0, so x = 5. Hence option A is correct. Option B has signs that make the sum negative, while options C and D do not have the required sum and product.

(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.

(x^2-36=x^2-6^2), so it is solved quickly by difference of squares. In exams, recognizing (a^2-b^2) saves time.

Taking the square root gives \(x=\pm\sqrt{64}=\pm 8\), so the two solutions are \(x=8\) and \(x=-8\). Writing only \(x=8\) or only \(x=-8\) is incomplete because both numbers have square 64. In exams, remember to include the \(\pm\) sign when taking the square root.

((x-7)=0) or ((x+1)=0), so (x=7) or (x=-1). In exams, set each factor equal to zero separately.

In \(x^2+12x+7=0\), the coefficient of \(x\) is 12. To form a perfect square, calculate \(\left(\frac{12}{2}\right)^2=6^2=36\), so 36 is added and subtracted. Therefore, option A is correct. Exam tip: for \(x^2+bx\), use \(\left(\frac{b}{2}\right)^2\). Taking 12 or 6 alone does not complete the square.

From standard form (ax^2+bx+c=0), (a=4), (b=-3), and (c=-1). In exams, write the signs of (b) and (c) carefully.

Dividing every term of 6x² − 24 = 0 by 6 gives x² − 4 = 0, since 24 ÷ 6 = 4. Option D is incorrect because it removes the coefficient from x² but does not divide 24 by 6. In an exam, divide every term of an equation by the same non-zero number.

Since (-7+3=-4) and (-7\times3=-21), ((x-7)(x+3)) is correct. In exams, choose mixed signs carefully.

In \(x^2+14x+49\), we have \(49=7^2\) and the middle term is \(14x=2\times7\times x\). Therefore, it is the perfect square \((x+7)^2\), so the equation can be written as \((x+7)^2=0\). Option B would produce the middle term \(-14x\). In exams, match the expression with \(a^2+2ab+b^2=(a+b)^2\).

Since \(x^2+14x+49=(x+7)^2\), the equation becomes \((x+7)^2=0\), giving \(x=-7\). This root occurs twice, so it is the repeated root. Exam tip: for \((x+a)^2=0\), the root is \(x=-a\), not \(x=a\).

The governing concept is extracting the greatest common factor from every term without changing the value of the expression. In 5x² + 15x, both terms contain 5x. Dividing each term by 5x leaves x in the first term and 3 in the second, because 5x² ÷ 5x = x and 15x ÷ 5x = 3. Thus 5x² + 15x = 5x(x + 3), and the equation becomes 5x(x + 3) = 0. Therefore option A is correct. Option B loses a factor x, option C changes the sign of the second term, and option D also incorrectly changes plus to minus.