Methods of Solving Quadratic Equations

(x^2-2x+5=(x-1)^2+4), so no real roots are obtained. In exams, the completed square form also shows the nature of roots.

Dividing the equation by 3 gives \(x^2=4\). Taking square roots, \(x=\pm\sqrt{4}=\pm2\), so the two roots are 2 and −2. Options B and C show only one of the two roots, while option D results from an incorrect square-root calculation. Exam tip: whenever \(x^2=a\), write \(x=\pm\sqrt{a}\) to include both roots.

The square-root property states that if u² = c, then u = ±√c, because both the positive and negative square roots have the same square. Here u = x − 5 and c = 3, so x − 5 = ±√3. Adding 5 to both sides gives x = 5 ± √3. Thus the two roots are 5 + √3 and 5 − √3, and option A is correct. Option B changes the sign of 5 incorrectly, option C replaces √3 by 3, and option D changes both the structure and the value. The ± sign is essential because a nonzero square has two real square roots.

(x^2+10x+21=(x+3)(x+7)), so (x=-3,-7). In exams, a positive middle term and positive constant can give both negative roots.

\((5x+1)(x+1)=5x^2+5x+x+1=5x^2+6x+1\), so option A is correct. Option D expands to \(5x^2+11x+6\), so it does not match the given equation. In an exam, multiply the factors back to verify them.

This quadratic can be solved efficiently by factorisation. We seek two factors whose product is 5x² + 6x + 1. The expression factors as (5x + 1)(x + 1), because multiplication gives 5x² + 5x + x + 1 = 5x² + 6x + 1. By the zero-product property, either 5x + 1 = 0 or x + 1 = 0. These give x = −1/5 and x = −1, respectively. Therefore option A is correct. Option B has the wrong signs, option C confuses a coefficient with the solution of 5x + 1 = 0, and option D makes both roots positive, which does not satisfy the original equation.

(x^2+x-1=0) has no simple integer factors, so the formula method is easier. In exams, the quadratic formula is safe in such cases.

Here (D=1-4(1)(-1)=5), so (x=\frac{-1\pm\sqrt{5}}{2}). In exams, keep the sign of (c=-1) correct.

((2x+1)(x-2)=2x^2-3x-2), so this is the correct factorised form. In exams, check the answer by expanding.

Factorise the expression as \\(2x^2-3x-2=(2x+1)(x-2)\\). Thus, \\(2x+1=0\\) gives \\(x=-\frac{1}{2}\\), and \\(x-2=0\\) gives \\(x=2\\). Therefore, option A is correct. Exam tip: Set each factor equal to zero and verify both roots in the original equation; option B has the signs reversed and is therefore incorrect.

From (x^2=25), (x=\pm\sqrt{25}=\pm5). In exams, both signs are necessary in the square root method.

Begin by moving the constant term: x² − 8x = −15. To complete the square, take half of the coefficient of x, which is −8/2 = −4, and square it: (−4)² = 16. Add 16 to both sides, obtaining x² − 8x + 16 = −15 + 16 = 1. The left side is the perfect square (x − 4)², so the correct middle step is (x − 4)² = 1. Hence option A is correct. Option B has the wrong sign, option C uses the coefficient incorrectly, and option D omits the 16 added to the right-hand side and therefore violates equality.

Rearranging the equation gives \(x^2-8x+16=1\), so \((x-4)^2=1\). Therefore, \(x-4=\pm1\), which gives \(x=5\) or \(x=3\). In option C, the sum of the roots is correct, but their product is \(7\), not the required \(15\). In exams, remember to consider both signs of \(\pm\) when completing the square.

(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 middle-term splitting for a quadratic expression ax² + bx + c. We need two numbers whose product is ac and whose sum is b. Here a = 8, b = 14, and c = 3, so ac = 8 × 3 = 24. The factor pair 12 and 2 has product 24 and sum 14. Therefore 14x can be written as 12x + 2x, giving 8x² + 12x + 2x + 3 = 0, which is option A. The other proposed pairs either do not add to 14 or do not multiply to 24. This split then permits grouping: 4x(2x + 3) + 1(2x + 3) = 0.

(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.

Here (D=(-8)^2-4(1)(12)=16), so (x=\frac{8\pm4}{2}). In exams, keep the sign of (-b) correct.

The governing concept is completing the square while preserving equality. Start with x² − 8x + 12 = 0 and move the constant term: x² − 8x = −12. Half of the coefficient of x is −4, and its square is 16. Add 16 to both sides, obtaining x² − 8x + 16 = 4. The left side is (x − 4)², so the correct middle step is (x − 4)² = 4, option A. The plus sign in option B would expand to x² + 8x + 16, not the required expression. Options C and D either use the wrong square or fail to balance the constant correctly.

(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.