Methods of Solving Quadratic Equations

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.

Here (D=(-10)^2-4(1)(24)=4), so (x=\frac{10\pm2}{2}). In exams, keep the sign of (-b) correct.

Adding (25) to (x^2-10x=-24) gives ((x-5)^2=1). In exams, add the same number to both sides.

(12x^2+17x+5=(12x+5)(x+1)), so the roots are (-\frac{5}{12}) and (-1). In exams, positive factors give negative roots.

Here (ac=-20) and (-20+1=-19), so the middle term is (-20x+x). In exams, check the sign of (ac) carefully.

(4x^2-19x-5=(4x+1)(x-5)), so the roots are (5) and (-\frac{1}{4}). In exams, reverse the signs from linear factors to write roots.

A quadratic equation is commonly written in standard form as ax² + bx + c = 0, with a ≠ 0. Starting from 5x² + 9x = 2, subtract 2 from both sides to preserve equality: 5x² + 9x − 2 = 2 − 2 = 0. Thus the standard-form equation is 5x² + 9x − 2 = 0, which is option A. Option B results from changing the sign incorrectly while moving 2. Option C changes the sign of the 9x term without justification, and option D interchanges the coefficients of x² and x. No solving of the quadratic is required; the task only asks for rearrangement into standard form.

(5x^2+9x-2=(5x-1)(x+2)), so the roots are (\frac{1}{5}) and (-2). In exams, correct standard form is very important.

Use the identity (p − q)² = p² − 2pq + q². Taking p=6x and q=5 gives (6x − 5)² = (6x)² − 2(6x)(5) + 5² = 36x² − 60x + 25. Therefore the equation 36x² − 60x + 25 = 0 is exactly equivalent to (6x − 5)² = 0, so option A is correct. The plus form would produce a positive middle term, 60x, not −60x. In option C, squaring 36x gives a leading term 1296x², and option D gives only x² as the leading term. The coefficients must therefore be matched before identifying the perfect-square identity.

((6x-5)^2=0), so (6x-5=0) and (x=\frac{5}{6}). In exams, write the repeated root as a correct fraction.

(8x^2-32x=8x(x-4)), so (x=0) and (x=4) are both roots. In exams, dividing by the variable can miss (x=0).

Comparing the equation with \(ax^2+bx+c=0\), we get \(a=5\), \(b=-10\), and \(c=-3\). Thus, \(D=b^2-4ac=(-10)^2-4(5)(-3)=100+60=160\). Since \(c\) is negative, the term \(-4ac\) becomes positive. Exam tip: write the sign of \(b\) explicitly before squaring it.

The formula gives (x=\frac{10\pm\sqrt{160}}{10}=1\pm\frac{2\sqrt{10}}{5}). In exams, simplify (\sqrt{160}=4\sqrt{10}).

For a quadratic equation ax² + bx + c = 0 to have equal real roots, its discriminant must be zero: D = b² − 4ac = 0. In x² − 16x + k = 0, a=1, b=−16, and c=k. Hence D = (−16)² − 4(1)(k) = 256 − 4k. Setting this equal to zero gives 256 − 4k = 0, so 4k=256 and k=64. Therefore option A is correct. The value 16 is the magnitude of the linear coefficient, not the required parameter. The value 32 comes from an incomplete division, while 256 ignores the term 4ak. The discriminant condition is the governing concept.

For equal roots, (k^2-144=0), so (k=\pm12), and (k>0) gives (k=12). In exams, apply the given condition.

The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.

Using \((a-b)^2=a^2-2ab+b^2\), we get \((7x-3)^2=49x^2-42x+9\). Hence the equation becomes \((7x-3)^2=0\). In option B, the middle term would be \(+42x\), while options C and D have an incorrect coefficient of \(x^2\). Exam tip: take the square roots of the first and last terms and verify the middle term using \(2ab\).

((7x-3)^2=0), so (7x-3=0) and (x=\frac{3}{7}). In exams, solve the linear equation after square form.