Methods of Solving Quadratic Equations

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). In this equation, 9x² is (3x)² and 16 is 4². Substituting A = 3x and B = 4 gives 9x² − 16 = (3x)² − 4² = (3x − 4)(3x + 4). Therefore the equation becomes (3x − 4)(3x + 4) = 0, so option A is correct. Option B expands to 9x² + 32x − 16, which contains an unwanted middle term. Option C represents a repeated factor and gives a different polynomial, while D also produces incorrect terms.

((3x-4)(3x+4)=0), so (x=\frac{4}{3}) or (x=-\frac{4}{3}). In exams, solve linear factors carefully.

(100=10^2), so it is a difference of squares form. In exams, choosing the correct identity is the fastest method.

(x^2+8x+12=(x+2)(x+6)), so the roots are (-2) and (-6). In exams, a positive middle term can give negative roots.

(x^2-5x=x(x-5)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.

(5+8=13) and (5\times8=40), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.