Methods of Solving Quadratic Equations

(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.

The direct answer is option A: x = 7 or x = −1. The zero-product rule says that if a product is zero, at least one factor must be zero. Thus (x − 7)(x + 1) = 0 gives two separate equations: x − 7 = 0 or x + 1 = 0. Solving them gives x = 7 and x = −1. Option A lists both roots correctly. Option B reverses both signs; substituting x = −7 does not make x − 7 zero, and x = 1 does not make x + 1 zero. Option C has x = 7 correct but changes −1 to 1, which fails the second factor. Option D changes 7 to −7, although x − 7 = 0 clearly requires x = 7. A direct substitution check confirms both answers: at 7, the first factor is zero; at −1, the second factor is zero. Common mistake: do not set both factors to zero simultaneously; use “or,” because either factor being zero makes the product zero.

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 algebraic principle is taking out a common factor from every term while preserving the original expression. Both terms, 5x² and 15x, contain 5x. Dividing the first term by 5x gives x, and dividing the second term by 5x gives 3. Therefore 5x² + 15x = 5x(x + 3), so the equation becomes 5x(x + 3) = 0. Expanding this expression returns 5x² + 15x, confirming the result. Option B omits the common factor x and therefore cannot reproduce the original quadratic term. Option C has an incorrect negative sign, while option D also changes the plus sign to minus. Thus option A is the unique correct form.

Factoring gives 5x² + 15x = 5x(x + 3) = 0. By the zero-product property, x = 0 or x + 3 = 0, so x = -3. Therefore, the roots are 0 and -3. Option B results from changing the sign of -3 incorrectly. In an exam, remember to include both roots.

(x^2-11x+28=(x-4)(x-7)), so the roots are (4) and (7). In exams, check (4+7=11) and (4\times7=28).

(6+(-4)=2) and (6\times(-4)=-24), 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 \(a=3\) and \(c=3\). Therefore, \(ac=3\times3=9\). The value 10 is \(b\), not \(ac\). In an exam, calculate \(ac\) before splitting the middle term.

(9+1=10) and (9\times1=9=ac), so (10x) is split as (9x+x). In exams, keep both sum and product correct.

From \(x^2-81=0\), we get \(x^2=81=9^2\). Hence, \(x=9\) or \(x=-9\), which is written as \(x=\pm9\). Options C and D give only one of the two valid roots, while option B incorrectly treats 81 as the square root of 81. Exam tip: use the difference of squares, \(a^2-b^2=(a-b)(a+b)\), to find both roots quickly.

Taking common factor (7x) gives (7x(x-2)=0). In exams, you can check by expanding after factoring.

Factor the equation as \(7x^2-14x=7x(x-2)=0\). By the zero-product property, \(7x=0\) or \(x-2=0\), giving the roots \(x=0\) and \(x=2\). In option B, the sign of the second root is incorrect. In an exam, first take out the common factor and then apply the zero-product rule.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=-2, c=1\), so \(D=(-2)^2-4(1)(1)=4-4=0\). Therefore, option A is correct. Choosing 4 may result from forgetting to subtract \(4ac\). Exam tip: always include the negative sign of \(b\) before squaring it.

(D=25>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. The expression x² + 4x + 3 has small integer coefficients, and 3 can be factored as 1 × 3 while 1 + 3 = 4. Therefore x² + 4x + 3 = (x + 1)(x + 3). The zero-product property then gives x = −1 or x = −3. Factorisation is direct and efficient here, so option A is correct. A graph could also display the roots, but it is not the easiest exact method. Long division is not the standard method for this task, and “table method” is not an appropriate algebraic procedure.

Factoring gives \(x^2+4x+3=(x+1)(x+3)\). Thus, \((x+1)(x+3)=0\) implies \(x=-1\) or \(x=-3\), so the roots are \(\{-1,-3\}\). Option B has the signs reversed; \(x+1=0\) gives \(x=-1\). Exam tip: Set each linear factor equal to zero to find the roots.

Half of (-18) is (-9), and ((-9)^2=81). In exams, the square of half the coefficient is always positive.

Taking square roots gives \(x=\pm\sqrt{121}=\pm 11\), so both \(x=11\) and \(x=-11\) are solutions. Writing only \(x=11\) is incomplete because \((-11)^2\) is also 121; \(\pm121\) results from not taking the square root. Exam tip: for \(x^2=a\) with \(a>0\), always write both roots as \(x=\pm\sqrt{a}\).

The governing identity is the difference of two squares: a² − b² = (a − b)(a + b). Rewrite 9x² − 16 as (3x)² − 4². Substituting a = 3x and b = 4 gives (3x − 4)(3x + 4), so the equation becomes (3x − 4)(3x + 4) = 0. Expanding verifies the result: 9x² + 12x − 12x − 16 = 9x² − 16, because the middle terms cancel. Option B expands to an expression containing an incorrect linear term, while option C represents a repeated factor and does not equal the original difference. Option D also gives incorrect coefficients. Hence option A is the correct factorisation.