Methods of Solving Quadratic Equations

(D=(-10)^2-4(1)(21)=16), so (x=\frac{10\pm4}{2}) gives (3) and (7). In exams, keep the sign of (b) correct in the formula.

(-8+(-10)=-18) and ((-8)(-10)=80), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.

(x^2-18x+80=(x-8)(x-10)), so the roots are (8) and (10). In exams, change signs while writing roots from factors.

From (9x^2-45x=0), (9x(x-5)=0), so (x=0) and (x=5). In exams, dividing by the variable can miss one root.

Completing the square means first moving the constant term and then adding the square of half the coefficient of x. From x² + 2x − 24 = 0, we obtain x² + 2x = 24. Half of the coefficient 2 is 1, and its square is 1. Adding 1 to both sides gives x² + 2x + 1 = 24 + 1, so (x + 1)² = 25. Thus option A is correct. Option B uses the wrong sign inside the square, while options C and D fail to add the required 1 to the right-hand side or use an incorrect constant. The same step would subsequently give x + 1 = ±5.

Factoring the quadratic gives \(x^2+2x-24=(x+6)(x-4)\). Thus, \((x+6)(x-4)=0\) gives \(x=-6\) or \(x=4\), so option A is correct. In option B, the signs of both roots are reversed. In an exam, set each factor equal to zero to obtain and check both roots.

(ac=-12) and (4+(-3)=1), so (x) is split as (4x-3x). In exams, check the sign of (ac) carefully.

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

The equal root is (x=\frac{-b}{2a}), so (x=\frac{12}{6}=2). In exams, this short formula is useful when (D=0).

Here (D=(-4)^2-4(1)(13)=-36<0), so there are no real roots. In exams, (D<0) means no real solution.

(x^2-4x+13=(x-2)^2+9), so no real roots are obtained. In exams, use completed square form to understand the nature of roots.

Dividing both sides by 5 gives \(x^2=16\). Taking square roots, \(x=\pm\sqrt{16}=\pm4\), so option A is correct. Writing only \(x=4\) or only \(x=-4\) omits one root, while \(\pm16\) results from an incorrect calculation. Exam tip: always include both the positive and negative signs when taking the square root of a positive number.

The governing idea is the square-root property: if u² = a, then u = ±√a. Taking u = x + 6 in the given equation gives x + 6 = ±√5. Subtracting 6 from both sides produces x = −6 ± √5. Therefore the two solutions are −6 + √5 and −6 − √5, and option A is correct. The plus-minus sign is essential because both a positive and a negative number have square 5. Option B incorrectly changes the sign of the constant when transposing, option C replaces √5 by 5, and option D changes both the radical expression and the algebraic arrangement.

(x^2+12x+32=(x+4)(x+8)), so (x=-4,-8). In exams, a positive middle term and positive constant can give negative roots.

Expanding (7x + 1)(x + 1) gives 7x² + 7x + x + 1 = 7x² + 8x + 1, so option A is correct. In option C, the coefficient of x² is 1, while option D expands to 7x² + 15x + 8. In exams, verify factorisation by expanding the factors and comparing the result with the original polynomial.

Here (D=2^2-4(1)(-2)=12), so (x=\frac{-2\pm\sqrt{12}}{2}=-1\pm\sqrt{3}). In exams, simplify (\sqrt{12}=2\sqrt{3}).

((3x+2)(x-4)=3x^2-10x-8), so this is the correct factorised form. In exams, check the answer by expanding.

The factorisation of the equation is \(3x^2-10x-8=(3x+2)(x-4)\). Thus, \((3x+2)(x-4)=0\) gives \(x=-\frac{2}{3}\) or \(x=4\). Therefore, option A is correct; both signs are incorrect in option B. Exam tip: after factorising, set each factor equal to zero to obtain the roots.

From (x^2=49), (x=\pm\sqrt{49}=\pm7). In exams, both signs are necessary in the square root method.

Adding (9) to (x^2-6x=7) gives ((x-3)^2=16). In exams, add the square of half the coefficient to both sides.