Methods of Solving Quadratic Equations

(x^2-x-30=(x-6)(x+5)), so the roots are (6) and (-5). In exams, the larger value decides the sign of the middle term.

(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.

The governing process is completing the square by adding the square of half the coefficient of x to both sides. From x² + 2x − 24 = 0, first move the constant term to get x² + 2x = 24. Half of the coefficient 2 is 1, and 1² = 1. Add 1 to both sides: x² + 2x + 1 = 24 + 1 = 25. The left side factors as (x + 1)², so the correct step is (x + 1)² = 25, option A. Option B has the wrong sign because its expansion contains −2x. Options C and D either use the wrong completed square or omit the required addition of 1 to the right side. The next step would be 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 principle is the square-root property: if u² = a, then u = ±√a. Let u = x + 6. From (x + 6)² = 5, we obtain x + 6 = ±√5. Subtracting 6 from both sides gives x = −6 ± √5. Thus the two solutions are x = −6 + √5 and x = −6 − √5, so option A is correct. The plus-minus sign is necessary because both positive and negative square roots have square 5. Option B changes the sign incorrectly when 6 is transposed, option C replaces √5 with 5, and option D changes both the radical and the algebraic arrangement. Substitution confirms that each value makes the square equal to 5.

(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}).

The direct answer is option A, (3x + 2)(x − 4) = 0. To verify a factorisation, multiply the factors. Expanding option A gives (3x + 2)(x − 4) = 3x^2 − 12x + 2x − 8 = 3x^2 − 10x − 8, exactly the given quadratic. Therefore option A is correct, and its roots would be x = −2/3 and x = 4. Option B expands to 3x^2 + 12x − 2x − 8 = 3x^2 + 10x − 8, so the middle sign is wrong. Option C expands to 3x^2 + 6x − 4x − 8 = 3x^2 + 2x − 8, which also has the wrong middle term. Option D expands to x^2 − 4x − 32, so even the coefficient of x^2 and the constant term do not match. The most reliable exam check is to multiply the proposed factors and compare all three terms: the x^2 term, the x term, and the constant term.

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.

From the equation, \(x^2-6x=7\). Adding 9 to both sides gives \(x^2-6x+9=16\), or \((x-3)^2=16\). Hence, \(x-3=\pm4\), so the roots are \(x=7\) and \(x=-1\). Option C is incorrect because its roots have a sum of 7, whereas the sum of the roots of this equation must be 6. Exam tip: In the completing-square method, use the \(\pm\) sign to obtain both roots.

(7x^2-9x+2=(7x-2)(x-1)), so the roots are (1) and (\frac{2}{7}). In exams, solve each linear factor separately.

Here (ac=30) and (-15+(-2)=-17), so the middle term is (-15x-2x). In exams, check both sum and product.

The correct answer is option A: x=4 and x=6. For x^2-10x+24=0, compare with ax^2+bx+c=0: a=1, b=-10, c=24. The discriminant is D=b^2-4ac=(-10)^2-4(1)(24)=100-96=4. The quadratic formula is x=\(\frac{-b\pm\sqrt D}{2a}\). Thus x=\(\frac{10\pm2}{2}\), giving x=6 or x=4. Option A is correct. Option B gives both roots negative and ignores the positive value of -b=10. Option C, 2 and 12, does not result from the formula; their sum is 14 rather than 10. Option D lists 5 and 24, but 24 is the constant term, not a root. A useful check is that the roots have sum 10 and product 24: 4+6=10 and 4×6=24. Exam cue: for b=-10, -b is +10.