Methods of Solving Quadratic Equations

(4x^2-12x+5=(2x-1)(2x-5)), so the roots are (\frac{1}{2}) and (\frac{5}{2}). In exams, solve each linear factor separately.

The governing method is splitting the middle term of ax² + bx + c into two terms whose coefficients add to b and whose product equals ac. Here a = 8, b = 14, and c = 3, so ac = 8 × 3 = 24. The numbers 12 and 2 have product 24 and sum 14. Therefore 14x can be replaced by 12x + 2x, giving 8x² + 12x + 2x + 3 = 0. This is option A. It also permits factorisation by grouping: 4x(2x + 3) + 1(2x + 3) = 0. Option B has coefficients whose product is 40, option C has product 48, and option D has the correct product but sum 14 only through 16 + (−2), though its grouping does not produce the required common factor pattern for this quadratic.

(8x^2+14x+3=(4x+1)(2x+3)), so the roots are (-\frac{1}{4}) and (-\frac{3}{2}). In exams, write fractional roots in simplest form.

The direct answer is option A, x = 2, 6. For x^2 − 8x + 12 = 0, compare with ax^2 + bx + c = 0: a = 1, b = −8, c = 12. The discriminant is D = b^2 − 4ac = (−8)^2 − 4(1)(12) = 64 − 48 = 16, so the square root of D is 4. The quadratic formula gives x = (−b ± √D)/(2a) = (8 ± 4)/2. With the plus sign, x = 12/2 = 6; with the minus sign, x = 4/2 = 2. Hence the roots are 2 and 6. Option A is correct. Option B, −2 and −6, incorrectly changes the signs; substituting either negative value does not satisfy the equation. Option C, 4 and 12, confuses the numerator values with the final roots. Option D, 3 and 5, does not result from the formula and their sum is 8, whereas the roots should sum to 8 but their product would be 15 instead of 12. Always use −b carefully.

The governing concept is completing the square while maintaining equality. Begin with x² − 8x + 12 = 0 and move the constant term to obtain x² − 8x = −12. Half of the coefficient of x, −8, is −4; its square is 16. Add 16 to both sides: x² − 8x + 16 = −12 + 16 = 4. The left side is the perfect square (x − 4)², so the correct middle step is (x − 4)² = 4, option A. Option B has the wrong sign because (x + 4)² produces +8x. Options C and D either use an incorrect square or fail to adjust the right-hand side after adding 16, violating equality.

(3x^2+8x+4=(3x+2)(x+2)), so the roots are (-\frac{2}{3}) and (-2). In exams, positive factors give negative roots.

Here (ac=-30) and (-15+2=-13), so the middle term is (-15x+2x). In exams, check the sign of (ac) carefully.

(5x^2-13x-6=(5x+2)(x-3)), so the roots are (3) and (-\frac{2}{5}). In exams, reverse the signs from linear factors to write roots.

The correct answer is option A: 3x^2+7x-6=0. Standard quadratic form is ax^2+bx+c=0, so the entire equation must first be rearranged to have zero on the right. Starting with 3x^2+7x=6, subtract 6 from both sides. This gives 3x^2+7x-6=0. Option A is exactly this result. Option B incorrectly keeps the constant as +6; moving +6 from the right to the left changes it to -6. Option C changes the sign of 7x without justification. Option D swaps the coefficients of x^2 and x, so it is a different equation. No division or factorisation is needed for this question. The safe method is to move the right-side term across the equality sign and reverse its sign. Memory cue: right-side positive constant becomes negative on the left.

(3x^2+7x-6=(3x-2)(x+3)), so the roots are (\frac{2}{3}) and (-3). In exams, correct standard form is necessary first.

((4x-3)^2=0), so (4x-3=0) and (x=\frac{3}{4}). In exams, write the repeated root as a correct fraction.

(6x^2-18x=6x(x-3)), so (x=0) and (x=3) are both roots. In exams, dividing by the variable can miss (x=0).

For ax² + bx + c = 0, the discriminant is D = b² − 4ac. Here a = 3, b = −6 and c = −2, so D = (−6)² − 4(3)(−2) = 36 + 24 = 60.

The formula gives (x=\frac{6\pm\sqrt{60}}{6}=1\pm\frac{\sqrt{15}}{3}). In exams, simplify (\sqrt{60}=2\sqrt{15}).

For equal roots, (k^2-100=0), so (k=\pm10), and (k<0) gives (k=-10). In exams, apply the given condition.

The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.

Using the identity \(a^2-2ab+b^2=(a-b)^2\), we get \(25x^2-20x+4=(5x)^2-2(5x)(2)+2^2=(5x-2)^2\). Hence, the correct perfect-square form is \((5x-2)^2=0\). In option B, the middle term would be \(+20x\), while options C and D have incorrect leading terms, \(625x^2\) and \(x^2\), respectively. Exam tip: Always check the sign and coefficient of the middle term when applying a perfect-square identity.

((5x-2)^2=0), so (5x-2=0) and (x=\frac{2}{5}). In exams, solve the linear equation after square form.

The standard form of a quadratic equation is ax² + bx + c = 0. Moving 7x to the left changes its sign to −7x, giving 2x² − 7x + 1 = 0. Therefore, option A is correct. Exam tip: the sign of a term changes whenever it is transposed to the other side of the equation.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=2, b=-7, c=1\\), so \\(D=(-7)^2-4(2)(1)=49-8=41\\). Option B results from taking only \\(b^2=49\\) and not subtracting \\(4ac\\). In an exam, identify \\(a,b,c\\) first and substitute their signed values carefully.

The correct answer is option A: (x-8)^2=36. Begin with x^2-16x+28=0. Move 28 to the right: x^2-16x=-28. To complete the square, take half of the coefficient of x, namely -16/2=-8, and square it: (-8)^2=64. Add 64 to both sides: x^2-16x+64=-28+64=36. The left side is (x-8)^2, so (x-8)^2=36. Option A is correct. Option B has the wrong sign inside the square; (x+8)^2 produces +16x, not -16x. Option C incorrectly uses 16 instead of half the coefficient and also mishandles the constant. Option D adds no 64, so the square is not completed. Memory cue: for x^2+bx, add (b/2)^2; keep the sign inside the bracket consistent with the middle term.

Factoring the equation gives \(x^2-16x+28=(x-2)(x-14)\). Thus, \(x-2=0\) or \(x-14=0\), so the roots are \(x=2,14\). In option D, the sum of the numbers is 14, whereas the sum of the roots must be 16. As an exam tip, verify the roots using their sum \(16\) and product \(28\).

By Vieta’s sum-of-roots relation, the sum of the roots is \(-p\). Here, \((-4)+(-5)=-9\), so \(-p=-9\) and hence \(p=9\). Option B results from incorrectly taking the root sum as \(p\) instead of \(-p\). In exams, remember that for \(x^2+px+q=0\), the sum of the roots is \(-p\).

If roots are (3) and (7), then ((x-3)(x-7)=0), that is (x^2-10x+21=0). In exams, form factors with opposite signs of roots.

(3x^2+11x+10=(3x+5)(x+2)), so the roots are (-\frac{5}{3}) and (-2). In exams, positive factors give negative roots.