Methods of Solving Quadratic Equations

To solve an equation containing a square, take both square roots because a real number can have either a positive or a negative square root. From (x−7)²=11, we obtain x−7=±√11. Adding 7 to both sides gives x=7±√11. Thus the two solutions are x=7+√11 and x=7−√11, making option A correct. Option B incorrectly changes the sign of 7 when it is transposed. Option C uses 11 instead of its square root, and option D places the square root on the wrong number and also omits the correct structure. The ± symbol is essential because both values square to 11.

(x^2+14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.

Expanding (11x+1)(x+1) gives 11x^2+11x+x+1=11x^2+12x+1, so option A is correct. Option C expands to x^2+12x+11, whose coefficient of x^2 is 1 rather than 11. In an exam, multiply the proposed factors to verify the original quadratic equation.

((11x+1)(x+1)=0), so (x=-\frac{1}{11}) and (-1). In exams, ((11x+1)=0) gives (-\frac{1}{11}).

Here (D=3^2-4(1)(-3)=21), so (x=\frac{-3\pm\sqrt{21}}{2}). In exams, keep the sign of (c=-3) correct.

((2x+1)(2x-7)=4x^2-12x-7), so this is the correct factorised form. In exams, check the answer by expanding.

((2x+1)(2x-7)=0), so (x=-\frac{1}{2}) and (\frac{7}{2}). In exams, change signs while writing roots.

From (x^2=81), (x=\pm\sqrt{81}=\pm9). In exams, both signs are necessary in the square root method.

Adding (144) to (x^2-24x=-108) gives ((x-12)^2=36). In exams, add the square of half the coefficient to both sides.

Writing the equation in completed-square form gives \((x-12)^2=36\). Therefore, \(x-12=\pm6\), so \(x=12+6=18\) or \(x=12-6=6\). Hence, the roots are \(6\) and \(18\). In an exam, remember to use the \(\pm\) sign to obtain both roots; option C contains the correct root \(6\) but \(12\) is not a root.

For equal roots, (D=0), so ((-2k)^2-4(1)(25)=0) gives (k=5). In exams, apply the condition (k>0).

(6x^2-7x-3=(3x+1)(2x-3)), so the roots are (-\frac{1}{3}) and (\frac{3}{2}). In exams, solve both linear factors separately.

Here (ac=72) and (-9+(-8)=-17), so the correct split is (-9x-8x). In exams, check both sum (b) and product (ac).

(12x^2-17x+6=(3x-2)(4x-3)), so the roots are (\frac{2}{3}) and (\frac{3}{4}). In exams, write fractional roots in simplest form.

For equal roots, (D=0), so (4(k+1)^2-4k^2=0) gives (2k+1=0). In exams, write (D) first in parameter questions.

The sum of roots is (2+4=6), and (\frac{p+3}{2}=6), so (p=9). In exams, use (-\frac{b}{a}) for sum of roots.

First we get \(x^2-\frac{10}{3}x+1=0\), then \(\left(x-\frac{5}{3}\right)^2=\frac{16}{9}\). In exams, divide by (a) first when \(a\neq1\).

(3x^2-10x+3=(3x-1)(x-3)), so the roots are (\frac{1}{3}) and (3). In exams, you may verify by completing square or factoring.

For real and equal roots, (D=0), so (36-4m=0) gives (m=9). In exams, remember equal roots mean (D=0).

The roots of the first equation are (2,\frac{1}{2}), and the roots of the second are (2,\frac{2}{3}). In exams, solve both equations separately for common root.