Methods of Solving Quadratic Equations

Taking square roots gives \(x=\pm\sqrt{169}=\pm13\), so both \(x=13\) and \(x=-13\) are solutions. Writing only \(x=13\) is incomplete because both the positive and negative numbers have square 169. In an exam, remember to write both values as \(x=\pm\sqrt{a}\) when \(x^2=a\).

(16x^2-25=(4x)^2-5^2), so ((4x-5)(4x+5)) is obtained. In exams, identify both squares first.

((4x-5)(4x+5)=0), so (x=\frac{5}{4}) or (x=-\frac{5}{4}). In exams, solve linear factors carefully.

(144=12^2), so it is a difference of squares form. In exams, choosing the correct identity is the fastest method.

(x^2+9x+20=(x+4)(x+5)), so the roots are (-4) and (-5). In exams, a positive middle term can give negative roots.

(x^2-11x=x(x-11)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.

The governing concept is that division by a variable can discard a possible zero value of that variable. First factor the equation without dividing: x² − 11x = x(x − 11), so x(x − 11) = 0. By the zero-product property, either x = 0 or x − 11 = 0, which gives x = 11. If we divide the original equation by x, we implicitly assume x ≠ 0; consequently, the solution x = 0 is lost, and only x = 11 remains in the reduced equation. Therefore option A is correct. Substitution confirms both original roots: 0² − 11(0) = 0 and 11² − 11(11) = 0. The safe method is to factor first and consider every factor separately.

(9+10=19) and (9\times10=90), so ((x-9)(x-10)) is correct. In exams, take both negative signs for a negative middle term.

((x-9)(x-10)=0), so (x=9) and (x=10). In exams, roots are obtained by taking opposite signs of factors.

((-9)+(-2)=-11) and ((-9)(-2)=18=ac), so (-9x-2x) is correct. In exams, split the middle term by checking (ac).

Expanding \((3x-2)(x-3)\) gives \(3x^2-9x-2x+6=3x^2-11x+6\), so option A is correct. Option C expands to \(3x^2-9x+6\), whose coefficient of \(x\) is incorrect. In an exam, expand the factors to check both the middle term and the constant term.

Factoring the quadratic gives \\(3x^2-11x+6=(3x-2)(x-3)\\). Therefore, \\(3x-2=0\\) gives \\(x=\frac{2}{3}\\), and \\(x-3=0\\) gives \\(x=3\\). Hence, option A is correct. Option D incorrectly uses \\(\frac{3}{2}\\) instead of \\(\frac{2}{3}\\). In an exam, set each factor equal to zero and solve for the roots separately.

Taking square roots gives \\(x-4=\\pm 5\\). Thus, \\(x-4=5\\) gives \\(x=9\\), while \\(x-4=-5\\) gives \\(x=-1\\). Option B results from handling the addition of 4 incorrectly. In exams, remember to include both \\(\\pm\\) cases when taking a square root.

The coefficient of \(x\) is 24, so its half is 12. To complete the square, we add the square of this half: \(\left(\frac{24}{2}\right)^2=12^2=144\). Therefore, 144 is added to both sides. Remember that 12 is the half of the coefficient, while 144 is the number added.

Use the perfect-square identity
(a+b)^2=a^2+2ab+b^2. Taking
a=x and
b=12 gives
(x+12)^2=x^2+2(x)(12)+12^2=x^2+24x+144. Therefore, option A is correct. Note that
(x-12)^2 would have the middle term
-24x, so it is not correct. Exam tip: take the square root of the constant term and verify twice its product with the variable term.

(11+(-5)=6) and (11\times(-5)=-55), so ((x+11)(x-5)) is correct. In exams, keep one sign positive and one negative.

((x+11)(x-5)=0), so (x=-11) and (x=5). In exams, the sign changes while finding roots from factors.

Dividing both sides of the equation by 12 gives \(\frac{12x^2}{12}=\frac{108}{12}\), so \(x^2=9\). Solving further gives \(x=\pm3\). In an exam, remember to perform the same operation on both sides of an equation.

Dividing both sides by 12 gives \(x^2=9\). Therefore, \(x=\pm\sqrt{9}=\pm3\), so the two solutions are \(x=3\) and \(x=-3\). Writing only \(x=3\) is incomplete because the square of \(-3\) is also 9. In exams, remember to include both the positive and negative square roots.

In the quadratic formula, the part inside the square root is (b^2-4ac). In exams, it is also called the discriminant (D).