Methods of Solving Quadratic Equations

Write the equation by completing the square: \(x^2-12x+20=(x-6)^2-16=0\). Thus, \((x-6)^2=16\), so \(x-6=\pm4\), giving \(x=2\) or \(x=10\). The values in option C have sum 12 but product 32, whereas the product of the roots must be 20. Exam tip: always include both values obtained from \(\pm\) in a quadratic equation.

The factors corresponding to the roots are \((x+3)(x+4)\). Expanding gives \(x^2+7x+12=0\), so \(p=7\). Option B incorrectly uses the sum of the roots, \(-7\), whereas the coefficient is its negative. Exam tip: for \(x^2+px+q=0\), the sum of the roots is \(-p\).

If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is (x^2-7x+10=0). In exams, form factors with opposite signs of roots.

(2x^2+7x+6=(2x+3)(x+2)), so (x=-\frac{3}{2}) and (-2). In exams, positive factors give negative roots.

(x^2-3x-10=(x-5)(x+2)), so the roots are (5) and (-2). In exams, in mixed signs the larger value decides the middle-term sign.

(D=(-6)^2-4(1)(8)=4), so (x=\frac{6\pm2}{2}) gives (2) and (4). In exams, keep the sign of (-b) correct.

(-5+(-9)=-14) and ((-5)(-9)=45), so this pair is correct. In exams, if (c) is positive and (b) is negative, both numbers may be negative.

From (7x^2-28x=0), (7x(x-4)=0), so (x=0) and (x=4). In exams, dividing by the variable can miss one root.

Completing the square requires preserving equality while converting the quadratic expression into a perfect square. First move the constant term: x² + 4x = 12. The coefficient of x is 4, so take half of it, 2, and add its square, 2² = 4, to both sides. This gives x² + 4x + 4 = 12 + 4 = 16. The left side factors as (x + 2)², so the correct middle step is (x + 2)² = 16. Therefore option A is correct. Option B has the wrong sign inside the square. Option C uses 4 instead of half the coefficient and also changes the equality improperly. Option D adds the square to the left but fails to add it to the right, violating the balance of the equation.

Factor the equation as \\(x^2+6x-2x-12=0\\), giving \\( (x+6)(x-2)=0 \\). Therefore, \\(x=-6\\) or \\(x=2\\), so option A is correct. In option B, the signs of both roots are incorrect. As an exam check, the roots should have sum \\(-4\\) and product \\(-12\\).

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

The equal root is (x=\frac{-b}{2a}), so (x=\frac{8}{4}=2). In exams, this short formula is useful when (D=0).

Here (D=(-2)^2-4(1)(5)=-16<0), so there are no real roots. In exams, (D<0) means no real solution.

(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 governing principle is the square-root property: if u² = c, then u = ±√c. The plus-or-minus sign is necessary because both a positive and a negative number can have the same nonzero square. Here u = x − 5 and c = 3, so x − 5 = ±√3. Adding 5 to both sides gives x = 5 ± √3. Consequently, the two real solutions are x = 5 + √3 and x = 5 − √3, so option A is correct. Option B changes the sign of 5 incorrectly. Option C replaces √3 with 3, which is not valid, and option D changes both the algebraic structure and the values. Substitution of either value into the original equation confirms the result.

(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.

(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.

Direct answer: Option A, \\( (2x+1)(x-2)=0 \\). To check a factorisation, multiply every term: \\(2x\cdot x=2x^2\\), \\(2x\cdot(-2)=-4x\\), \\(1\cdot x=x\\), and \\(1\cdot(-2)=-2\\). Combining the middle terms gives \\(2x^2-4x+x-2=2x^2-3x-2\\), exactly the original equation. Option A is therefore correct. Option B gives \\(2x^2+4x-x-2=2x^2+3x-2\\), with the wrong sign of the middle term. Option C gives \\(2x^2+2x-2x-2=2x^2-2\\), missing the required linear term. Option D gives \\(x^2-4\\), which has the wrong leading coefficient and constant. Exam cue: multiply the two brackets and check all three terms.

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.

To complete the square, first isolate the variable terms by moving 15 to the right: x² − 8x = −15. The coefficient of x is −8, and half of it is −4. Squaring this half gives (−4)² = 16. Add 16 to both sides: x² − 8x + 16 = −15 + 16 = 1. The left side is the perfect square (x − 4)², because (x − 4)² expands to x² − 8x + 16. Therefore the correct middle step is (x − 4)² = 1, so option A is correct. Option B has the wrong sign, option C does not complete the square, and option D fails to add 16 to the right side.

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.