Methods of Solving Quadratic Equations

The governing rule for completing the square is to add and subtract (b/2)² when the coefficient of x² is 1. In x² + 20x + 13 = 0, the coefficient b is 20. Half of 20 is 10, and squaring it gives 10² = 100. Therefore 100 must be both added and subtracted so that the first three terms can form a perfect square: x² + 20x + 100 = (x + 10)². The equation may then be written as (x + 10)² − 100 + 13 = 0, or (x + 10)² = 87. Hence option A is correct. The number 20 is the linear coefficient, 10 is only its half, and 13 is the constant term; none of these is the required added square.

The governing concept is the standard form of a quadratic equation, ax² + bx + c = 0. To identify the coefficients, compare each term directly with that form. In 5x² + 2x − 7 = 0, the coefficient of x² is a = 5, the coefficient of x is b = 2, and the constant term is c = −7. The negative sign belongs to the constant term and must not be omitted. Therefore option A is correct. Option B swaps the coefficients of x² and x. Option C changes both signs of the linear and constant terms, and option D assigns the constant term as a and the other coefficients incorrectly. Correct identification is essential before substituting into the quadratic formula.

Dividing both sides by (10) gives (x^2-9=0). In exams, simplifying the equation first saves time.

Since (-9+3=-6) and (-9\times3=-27), ((x-9)(x+3)) is correct. In exams, choose mixed signs carefully.

Using \((a+b)^2=a^2+2ab+b^2\), we get \((x+9)^2=x^2+18x+81\), since \(2\times 9=18\) and \(9^2=81\). Therefore, the correct form is \((x+9)^2=0\). The closest distractor, \((x-9)^2\), has the middle term \(-18x\), not \(+18x\). Exam tip: for a perfect square, compare the square root of the constant term with half the coefficient of \(x\).

((x+9)^2=0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.

The governing concept is common-factor extraction. Both terms, 4x² and 28x, contain 4x. Dividing the first term by 4x gives x, and dividing the second term by 4x gives 7, because 28x ÷ 4x = 7. Since the original terms have a plus sign, the factor inside the bracket is x + 7. Thus 4x² + 28x = 4x(x + 7), and the equation becomes 4x(x + 7) = 0. Option A is correct. Option B omits the factor x and therefore does not expand to the original expression. Option C changes the sign and option D also incorrectly replaces the plus sign with a minus sign.

Factoring the equation gives \(4x^2+28x=4x(x+7)=0\). By the zero-product property, \(4x=0\) gives \(x=0\), and \(x+7=0\) gives \(x=-7\). Therefore, option A is correct. In option B, the sign of \(-7\) is incorrect. In an exam, remember to set both factors equal to zero after factoring.

(x^2-17x+72=(x-8)(x-9)), so the roots are (8) and (9). In exams, check (8+9=17) and (8\times9=72).

(8+(-5)=3) and (8\times(-5)=-40), so this pair is correct. In exams, match the sum with (b) and product with (c).

Comparing the equation with the standard form \(ax^2+bx+c=0\), we get \(a=4\) and \(c=3\). Therefore, \(ac=4\times3=12\). The value 13 is \(b\), while 3 is only \(c\). Exam tip: find \(ac\) before splitting the middle term.

(12+1=13) and (12\times1=12=ac), so (13x) is split as (12x+x). In exams, keep both sum and product correct.

Rewrite \(x^2-121=0\) as \(x^2-11^2=0\). Then \((x-11)(x+11)=0\), so \(x=11\) or \(x=-11\), that is, \(x=\pm 11\). Option B is incorrect because 121 is the value of \(x^2\), not of \(x\). In an exam, recognize \(121=11^2\) and use the difference-of-squares method.

The governing concept is taking the greatest common factor from every term of a polynomial. In 9x² − 27x, both terms contain 9x. Dividing each term by 9x gives x and −3 respectively, so the expression becomes 9x(x − 3). Therefore the factored equation is 9x(x − 3) = 0, making option A correct. A quick expansion verifies it: 9x multiplied by x gives 9x², and 9x multiplied by −3 gives −27x. Option B incorrectly omits the factor x, option C changes the sign of the constant term, and option D also has the wrong sign. The factor form can then be used with the zero-product property to find x = 0 or x = 3.

Factoring the equation gives \(9x^2-27x=9x(x-3)=0\). By the zero-product rule, \(x=0\) or \(x-3=0\), which gives \(x=3\). Therefore, the roots are \(0\) and \(3\). Option B is incorrect because the second root is positive \(3\), not \(-3\). In an exam, factor out the common term first and then apply the zero-product rule.

In the standard form \(ax^2+bx+c=0\), \(a=1\), \(b=2\), and \(c=1\). Therefore, \(D=b^2-4ac=2^2-4(1)(1)=4-4=0\), so option A is correct. Option D results from taking only \(b^2=4\) and not subtracting \(4ac\). Exam tip: When \(D=0\), the quadratic equation has equal roots.

(D=36>0), so two distinct real roots are obtained. In exams, connect (D>0) with distinct real roots.

The governing idea is to select a solution method that matches the structure and coefficients of the quadratic. Here the constant term 10 has factor pairs 1 and 10 or 2 and 5. The pair 5 and 2 has sum 7, so x² + 7x + 10 can be written as (x + 5)(x + 2) = 0. The zero-product property then gives x = −5 or x = −2. Thus factorisation is the easiest and most direct method, making option A correct. Long division is not normally required for solving this quadratic, and a table or graph could be used in some contexts but would be less direct and less exact for this simple expression. The small integer coefficients make factorisation especially convenient.

The factorisation of \(x^2+7x+10\) is \((x+5)(x+2)\). Therefore, \((x+5)(x+2)=0\) gives \(x=-5\) or \(x=-2\). Option B has the wrong signs; from \(x+a=0\), we get \(x=-a\). Exam tip: check that the constant terms multiply to 10 and add to 7.

Half of (-22) is (-11), and ((-11)^2=121). In exams, the square of half the coefficient is always positive.