Methods of Solving Quadratic Equations

From (11x^2-77x=0), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.

Direct answer: Option A, \\( (x+4)^2=49 \\). Begin with \\(x^2+8x-33=0\\), then move the constant: \\(x^2+8x=33\\). Half of the coefficient 8 is 4, and \\(4^2=16\\). Add 16 to both sides: \\(x^2+8x+16=33+16\\). The left side becomes \\( (x+4)^2\\), and the right side becomes 49. Therefore A is correct. B has the wrong sign: \\( (x-4)^2\\) contains \\(-8x\\), not +8x. C uses 8 instead of half of 8 and also leaves the right side incorrectly as 33. D uses the correct square form but forgets to add 16 to the right side. Exam cue: for \\(x^2+bx\\), add \\( (b/2)^2\\) to both sides.

The equation factors as \(x^2+8x-33=(x-3)(x+11)=0\). Therefore, \(x=3\) or \(x=-11\), so option A is correct. In option B, the signs of both roots are incorrect. As an exam check, the sum of the roots should be \(-8\) and their product should be \(-33\).

(ac=-24) and (4+(-6)=-2), so (-2x) is split as (4x-6x). In exams, check the sign of (ac) carefully.

(8x^2-2x-3=(4x-3)(2x+1)), so the roots are (\frac{3}{4}) and (-\frac{1}{2}). In exams, solve both linear factors carefully.

The equal root is (x=\frac{-b}{2a}), so (x=\frac{16}{8}=2). In exams, this short formula is useful when (D=0).

Here (D=(-6)^2-4(1)(18)=-36<0), so there are no real roots. In exams, (D<0) means no real solution.

(x^2-6x+18=(x-3)^2+9), so no real roots are obtained. In exams, use completed square form to understand the nature of roots.

Dividing both sides by 7 gives \\(x^2=25\\). Therefore, \\(x=\pm\sqrt{25}=\pm5\\), so the roots are 5 and -5. Writing only \\(x=5\\) or only \\(x=-5\\) is incomplete because both square roots must be considered. Exam tip: for \\(x^2=a\\), write \\(x=\pm\sqrt{a}\\).

(x^2+14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.

Expanding (11x+1)(x+1) gives 11x^2+11x+x+1=11x^2+12x+1, so option A is correct. Option C expands to x^2+12x+11, whose coefficient of x^2 is 1 rather than 11. In an exam, multiply the proposed factors to verify the original quadratic equation.

The direct answer is option A: the roots are \(x=-\frac{1}{11}\) and \(x=-1\). To solve, factor the quadratic. We need two factors whose product is \(11x^2+12x+1\). The factorisation is \((11x+1)(x+1)=0\), because multiplying gives \(11x^2+11x+x+1=11x^2+12x+1\). By the zero-product rule, either \(11x+1=0\), giving \(x=-\frac{1}{11}\), or \(x+1=0\), giving \(x=-1\). Option A is correct because it contains exactly these two values. Option B has positive values, but both roots must be negative because the factors contain plus signs. Option C gives \(-11\) instead of \(-\frac{1}{11}\); it confuses a coefficient with its reciprocal. Option D has positive values and is therefore incorrect. A useful check is that the sum of roots is \(-12/11\), which equals \(-1/11-1\), and their product is \(1/11\). Memory cue: after factoring, set each factor equal to zero and change the sign carefully.

Here (D=3^2-4(1)(-3)=21), so (x=\frac{-3\pm\sqrt{21}}{2}). In exams, keep the sign of (c=-3) correct.

The correct answer is option A: (2x+1)(2x-7)=0. Expand it carefully: (2x+1)(2x-7)=4x^2-14x+2x-7=4x^2-12x-7, exactly the given quadratic. Therefore option A is correct. Option B expands to (2x-1)(2x+7)=4x^2+14x-2x-7=4x^2+12x-7, so the middle term has the wrong sign. Option C expands to (4x-7)(x+1)=4x^2+4x-7x-7=4x^2-3x-7, not the required middle term. Option D expands to (x+7)(4x-1)=4x^2+28x-x-7=4x^2+27x-7, which is also different. Factoring can always be checked by multiplication. The first terms multiply to 4x^2 and the last terms multiply to -7; the cross terms must combine to -12x. Memory cue: expand every proposed factorisation before accepting it.

((2x+1)(2x-7)=0), so (x=-\frac{1}{2}) and (\frac{7}{2}). In exams, change signs while writing roots.

From (x^2=81), (x=\pm\sqrt{81}=\pm9). In exams, both signs are necessary in the square root method.

Adding (144) to (x^2-24x=-108) gives ((x-12)^2=36). In exams, add the square of half the coefficient to both sides.

Writing the equation in completed-square form gives \((x-12)^2=36\). Therefore, \(x-12=\pm6\), so \(x=12+6=18\) or \(x=12-6=6\). Hence, the roots are \(6\) and \(18\). In an exam, remember to use the \(\pm\) sign to obtain both roots; option C contains the correct root \(6\) but \(12\) is not a root.

For equal roots, (D=0), so ((-2k)^2-4(1)(25)=0) gives (k=5). In exams, apply the condition (k>0).

Begin by moving the constant term to the right: x² + 4x = −1. To complete the square, add (4/2)² = 2² = 4 to both sides. The left side becomes x² + 4x + 4 = (x+2)², while the right side becomes −1+4=3. Therefore the equivalent equation is (x+2)²=3, so option A is correct. Option B has the wrong sign inside the square and would create −4x. Option C adds an incorrect amount and has the wrong coefficient structure, while option D fails to account for the added 4 on the right. Adding the same quantity to both sides preserves equivalence.

The governing concept is solving a quadratic equation by recognizing a perfect square. Compare x² − 2√5x + 5 with the identity (x − c)² = x² − 2cx + c². Taking c = √5 gives c² = 5, so the equation becomes x² − 2√5x + 5 = (x − √5)² = 0. A square can equal zero only when its base is zero; hence x − √5 = 0 and x = √5. The root is repeated because both quadratic roots coincide. Therefore option A is correct. Option B has the wrong sign, while C and D confuse the constant 5 with the root and do not satisfy the equation.

The governing concept is completing the square while preserving equality. Start with x² + 6x + 2 = 0 and move the constant term to the other side: x² + 6x = −2. Half the coefficient of x is 6/2 = 3, and its square is 9. Add 9 to both sides, not just one side: x² + 6x + 9 = −2 + 9. The left side becomes (x + 3)² and the right side becomes 7, so (x + 3)² = 7. Therefore option A is correct. Option B has the wrong sign inside the square, option C uses 6 instead of half of 6, and option D adds 9 incorrectly or fails to account for the moved constant.

The governing method is completing the square, which adds the same quantity to both sides so that the quadratic expression becomes a perfect square. Starting with x^2+8x+5=0, move 5 to the right: x^2+8x=-5. Half of the coefficient of x is 8/2=4, and its square is 16. Add 16 to both sides: x^2+8x+16=-5+16=11. The left side factors as (x+4)^2, giving (x+4)^2=11. Thus option A is correct. Option B has the wrong sign because (x-4)^2 produces -8x; option C uses an incorrect square term, and option D omits the required change of 16 on the right-hand side.