Methods of Solving Quadratic Equations

In standard form, all terms are brought to one side and the other side is made zero. Subtracting \(11x\) from both sides gives \(3x^2+2-11x=0\), which can be written as \(3x^2-11x+2=0\). Therefore, option A is correct. Option B has the wrong sign for the linear term. Exam tip: a term changes its sign when it is transposed to the other side.

In the standard form \(ax^2+bx+c=0\), we have \(a=3\), \(b=-11\), and \(c=2\). Thus, the discriminant is \(D=b^2-4ac=(-11)^2-4(3)(2)=121-24=97\). The value 121 results from failing to subtract \(4ac\). In an exam, identify \(a,b,c\) first and then apply \(D=b^2-4ac\).

Adding (81) to (x^2-18x=-45) gives ((x-9)^2=36). In exams, add the square of half the coefficient.

The equation factors as \((x-3)(x-15)=0\). Therefore, \(x=3\) or \(x=15\), so option A is correct. In option C, the numbers add to 18 but their product is 72, whereas the required product is 45. In an exam, verify the roots using their sum 18 and product 45.

The given roots form the factors \((x+6)(x+7)=0\). Expanding gives \(x^2+13x+42=0\), so \(p=13\). The distractor \(-13\) results from missing the negative sign in the relation that the sum of roots is \(-p\): \(-6+(-7)=-13=-p\). Exam tip: For \(x^2+px+q=0\), the sum of the roots is \(-p\).

If roots are (4) and (9), then ((x-4)(x-9)=0), that is (x^2-13x+36=0). In exams, form factors with opposite signs of roots.

(5x^2+16x+3=(5x+1)(x+3)), so the roots are (-\frac{1}{5}) and (-3). In exams, positive factors give negative roots.

(x^2-4x-32=(x-8)(x+4)), so the roots are (8) and (-4). In exams, the larger value decides the sign of the middle term.

(D=(-14)^2-4(1)(45)=16), so (x=\frac{14\pm4}{2}) gives (5) and (9). In exams, keep the sign of (b) correct in the formula.

(-8+(-12)=-20) and ((-8)(-12)=96), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.

(x^2-20x+96=(x-8)(x-12)), so the roots are (8) and (12). In exams, change signs while writing roots from factors.

From (11x^2-77x=0), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.

Adding (16) to (x^2+8x=33) gives ((x+4)^2=49). In exams, add the square of half the coefficient.

The equation factors as \(x^2+8x-33=(x-3)(x+11)=0\). Therefore, \(x=3\) or \(x=-11\), so option A is correct. In option B, the signs of both roots are incorrect. As an exam check, the sum of the roots should be \(-8\) and their product should be \(-33\).

(ac=-24) and (4+(-6)=-2), so (-2x) is split as (4x-6x). In exams, check the sign of (ac) carefully.

(8x^2-2x-3=(4x-3)(2x+1)), so the roots are (\frac{3}{4}) and (-\frac{1}{2}). In exams, solve both linear factors carefully.

The equal root is (x=\frac{-b}{2a}), so (x=\frac{16}{8}=2). In exams, this short formula is useful when (D=0).

Here (D=(-6)^2-4(1)(18)=-36<0), so there are no real roots. In exams, (D<0) means no real solution.

(x^2-6x+18=(x-3)^2+9), so no real roots are obtained. In exams, use completed square form to understand the nature of roots.

Dividing both sides by 7 gives \\(x^2=25\\). Therefore, \\(x=\pm\sqrt{25}=\pm5\\), so the roots are 5 and -5. Writing only \\(x=5\\) or only \\(x=-5\\) is incomplete because both square roots must be considered. Exam tip: for \\(x^2=a\\), write \\(x=\pm\sqrt{a}\\).