Methods of Solving Quadratic Equations

Direct answer: Option A, \\( (x-5)^2=1 \\). Start with \\(x^2-10x+24=0\\). Move the constant term: \\(x^2-10x=-24\\). Half the coefficient of x is \\(-10/2=-5\\), and its square is 25. Add 25 to both sides: \\(x^2-10x+25=-24+25\\). The left side is \\( (x-5)^2\\), and the right side is 1. Thus A is correct. B has the wrong sign and would represent \\(x^2+10x+25\\). C wrongly uses \\(x-10\\) and ignores the half-coefficient rule. D leaves the right side as 24 and forgets the added 25. Option A is the exact middle step; later, it gives \\(x=4\\) or \\(x=6\\). Memory cue: add the square of half the x-coefficient 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.

(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.

((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}).

The governing concept is the discriminant condition for equal roots. A quadratic equation ax² + bx + c = 0 has equal real roots precisely when its discriminant D = b² − 4ac is zero. In x² − 16x + k = 0, the coefficients are a = 1, b = −16, and c = k. Therefore D = (−16)² − 4(1)(k) = 256 − 4k. Setting D equal to zero gives 256 − 4k = 0, so 4k = 256 and k = 64. Hence option A is correct. The value 16 is only the magnitude of the linear coefficient, not k. The value 32 reflects incomplete division, while 256 ignores the factor 4ac. With k = 64, the equation becomes (x − 8)² = 0, confirming equal roots.

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.

In standard form, all terms are brought to one side and the other side is made zero. Subtracting \(11x\) from both sides gives \(3x^2+2-11x=0\), which can be written as \(3x^2-11x+2=0\). Therefore, option A is correct. Option B has the wrong sign for the linear term. Exam tip: a term changes its sign when it is transposed to the other side.

In the standard form \(ax^2+bx+c=0\), we have \(a=3\), \(b=-11\), and \(c=2\). Thus, the discriminant is \(D=b^2-4ac=(-11)^2-4(3)(2)=121-24=97\). The value 121 results from failing to subtract \(4ac\). In an exam, identify \(a,b,c\) first and then apply \(D=b^2-4ac\).

The correct answer is option A: (x-9)^2=36. Start with x^2-18x+45=0 and move 45 to the right: x^2-18x=-45. Half of the coefficient of x is -18/2=-9, and its square is 81. Add 81 to both sides: x^2-18x+81=-45+81=36. The left side becomes (x-9)^2, so the required middle step is (x-9)^2=36. Option A is correct. Option B has (x+9)^2, whose middle term would be +18x, not -18x. Option C uses 18 instead of half of 18 and does not complete the square. Option D leaves the right side as 45, although the original 45 was moved and then must be adjusted after adding 81. Memory cue: half the x-coefficient, square it, and add it to both sides.

The equation factors as \((x-3)(x-15)=0\). Therefore, \(x=3\) or \(x=15\), so option A is correct. In option C, the numbers add to 18 but their product is 72, whereas the required product is 45. In an exam, verify the roots using their sum 18 and product 45.

The given roots form the factors \((x+6)(x+7)=0\). Expanding gives \(x^2+13x+42=0\), so \(p=13\). The distractor \(-13\) results from missing the negative sign in the relation that the sum of roots is \(-p\): \(-6+(-7)=-13=-p\). Exam tip: For \(x^2+px+q=0\), the sum of the roots is \(-p\).

If roots are (4) and (9), then ((x-4)(x-9)=0), that is (x^2-13x+36=0). In exams, form factors with opposite signs of roots.

(5x^2+16x+3=(5x+1)(x+3)), so the roots are (-\frac{1}{5}) and (-3). In exams, positive factors give negative roots.

(x^2-4x-32=(x-8)(x+4)), so the roots are (8) and (-4). In exams, the larger value decides the sign of the middle term.

(D=(-14)^2-4(1)(45)=16), so (x=\frac{14\pm4}{2}) gives (5) and (9). In exams, keep the sign of (b) correct in the formula.

(-8+(-12)=-20) and ((-8)(-12)=96), so this pair is correct. In exams, when (c) is positive and (b) is negative, both numbers are negative.

(x^2-20x+96=(x-8)(x-12)), so the roots are (8) and (12). In exams, change signs while writing roots from factors.