Methods of Solving Quadratic Equations

(18x^2-45x+14=(3x-1)(6x-14)), so the roots are (\frac{1}{3}) and (\frac{7}{3}). In exams, always check the final factors when a common factor appears.

Here (ac=900) and (-36+(-25)=-61), so the correct split is (-36x-25x). In exams, even for large (ac), match both sum and product.

(30x^2-61x+30=(5x-6)(6x-5)), so the roots are (\frac{6}{5}) and (\frac{5}{6}). In exams, keep the denominators of fractional roots correct.

For equal roots, (D=0), so ((k-5)^2=k^2-36) and (k=\frac{61}{10}). In exams, handle constant terms carefully while expanding squares.

The sum of roots is (4), and (\frac{p+6}{7}=4), so (p=22). In exams, use (-\frac{b}{a}) for the sum.

First (x^2-4x+\frac{9}{13}=0) is obtained, then ((x-2)^2=\frac{43}{13}). In exams, divide by (a) first when (a\neq1).

Since ((x-2)^2=\frac{43}{13}), (x=2\pm\sqrt{\frac{43}{13}}=2\pm\frac{\sqrt{559}}{13}). In exams, rationalize the denominator.

For real and equal roots, (D=0), so (576-4m=0) gives (m=144). In exams, equal roots indicate (D=0).

The first equation has roots (\frac{5}{2},\frac{8}{5}), and the second has roots (\frac{5}{2},\frac{4}{5}). In exams, solve both equations separately for the common root.

The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.

The other root is (15), so (q=10\times15=150). In exams, when (a=1), the constant term is the product of roots.

(12x^2+x-6=(3x-2)(4x+3)), so (x=\frac{2}{3},-\frac{3}{4}) is correct. In exams, change signs carefully from factors.

Since (17=(\sqrt{17})^2) and the middle term is (-2\sqrt{17}x), it is ((x-\sqrt{17})^2). In exams, identify perfect squares even with irrational coefficients.

Direct answer: Option A, x=√17, and it is a repeated root. Compare the equation with the square
(x-√17)^2
. Expanding gives x^2-2√17x+(√17)^2=x^2-2√17x+17, exactly the given equation. Hence (x-√17)^2=0. A square is zero only when its expression is zero, so x-√17=0 and x=√17. The same value occurs twice as the two roots. Option A is correct. Option B, -√17, has the wrong sign; substitution does not make the square zero. Option C, 17, is not the value that makes x-√17 zero. Option D, -17, also has the wrong sign and magnitude. The discriminant is zero as another check: b^2-4ac=68-68=0. Memory cue: recognize x^2-2ax+a^2=(x-a)^2.

The equation is equivalent to ((x-p)(x-q)=0), so the roots are (p) and (q). In exams, apply zero product rule to symbolic factors too.

It is ((x-u)^2-v^2=0), so (x-u=\pm v) and (x=u\pm v). In exams, quickly recognize the difference of squares.

Dividing the whole equation by (49) gives (x^2-(r+s)x+rs=0). In exams, removing the common factor first shortens the solution.

Since (x^4=(x^2)^2=y^2), the new equation is (y^2-25y+144=0). In exams, use substitution to form a quadratic.

From (y^2-25y+144=0), (y=9,16), so (x^2=9,16) and (x=\pm3,\pm4). In exams, do not forget to return to (x).

Multiplying both sides by (7x) gives (7+7x^2=50x), that is (7x^2-50x+7=0). In exams, remember the condition (x\neq0).

(7x^2-50x+7=(7x-1)(x-7)), so (x=\frac{1}{7}) and (7). In exams, check whether obtained roots are valid in the original equation.

Cross multiplication gives ((x+6)^2=49x), so (x^2+12x+36-49x=0), and (x^2-37x+36=0). In exams, cross multiply carefully.

(x^2-37x+36=(x-1)(x-36)), so (x=1) and (x=36). In exams, check solutions against excluded denominator values.

(D=(-14)^2-4(1)(13)=144), so (x=\frac{14\pm12}{2}) gives (1) and (13). In exams, if (D) is a perfect square, simplify quickly.

The sum of roots is (-\frac{p}{9}), so (-\frac{p}{9}=-10) gives (p=90). In exams, remember the sum formula (-\frac{b}{a}).