Methods of Solving Quadratic Equations

(16x^2-38x+15=(8x-3)(2x-5)), so the roots are (\frac{3}{8}) and (\frac{5}{2}). In exams, do not invert fractional roots.

Here (ac=600) and (-30+(-20)=-50), so the correct split is (-30x-20x). In exams, even for large (ac), match both sum and product.

(24x^2-50x+25=(6x-5)(4x-5)), so the roots are (\frac{5}{6}) and (\frac{5}{4}). In exams, keep the denominator coefficients correctly.

For equal roots, (D=0), so ((k-4)^2=k^2-25) and (k=\frac{41}{8}). In exams, handle constant terms carefully while expanding squares.

The sum of roots is (4), and (\frac{p+5}{6}=4), so (p=19). In exams, use (-\frac{b}{a}) for the sum.

First (x^2-4x+\frac{7}{11}=0) is obtained, then ((x-2)^2=\frac{37}{11}). In exams, divide by (a) first when (a\neq1).

Since ((x-2)^2=\frac{37}{11}), (x=2\pm\sqrt{\frac{37}{11}}=2\pm\frac{\sqrt{407}}{11}). In exams, rationalize the denominator.

For real and equal roots, (D=0), so (484-4m=0) gives (m=121). In exams, equal roots indicate (D=0).

The first equation has roots (\frac{3}{2},\frac{9}{4}), and the second has roots (\frac{3}{2},\frac{10}{9}). In exams, solve both equations separately for the common root.

The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.

The other root is (14), so (q=9\times14=126). In exams, when (a=1), the constant term is the product of roots.

(10x^2+x-3=(5x+3)(2x-1)), so (x=\frac{1}{2},-\frac{3}{5}) is correct. In exams, change signs carefully from factors.

Since (13=(\sqrt{13})^2) and the middle term is (2\sqrt{13}x), it is ((x+\sqrt{13})^2). In exams, identify perfect squares even with irrational coefficients.

The direct answer is option A, x=-√13. The equation is x^2+2√13x+13=0. Compare it with the identity (x+a)^2=x^2+2ax+a^2. Taking a=√13 gives (x+√13)^2=x^2+2√13x+(√13)^2=x^2+2√13x+13. Therefore the equation becomes (x+√13)^2=0. A square is zero only when its inside is zero, so x+√13=0 and x=-√13. This is a repeated root, meaning both roots have the same value. Option B, x=√13, has the wrong sign; substituting it gives a positive nonzero expression. Option C, x=-13, is not the value obtained from the square factor and has the wrong magnitude. Option D, x=13, has both wrong sign and magnitude. Memory cue: (x+a)^2=0 always gives x=-a.

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

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

Dividing the whole equation by (36) gives (x^2-(m+n)x+mn=0). In exams, removing the common factor first shortens the solution.

Since (x^4=(x^2)^2=y^2), the new equation is (y^2-20y+64=0). In exams, use substitution to form a quadratic.

From (y^2-20y+64=0), (y=4,16), so (x^2=4,16) and (x=\pm2,\pm4). In exams, do not forget to return to (x).

Multiplying both sides by (6x) gives (6+6x^2=37x), that is (6x^2-37x+6=0). In exams, remember the condition (x\neq0).

(6x^2-37x+6=(6x-1)(x-6)), so (x=\frac{1}{6}) and (6). In exams, check whether obtained roots are valid in the original equation.

Cross multiplication gives ((x+5)^2=36x), so (x^2+10x+25-36x=0), and (x^2-26x+25=0). In exams, cross multiply carefully.

(x^2-26x+25=(x-1)(x-25)), so (x=1) and (x=25). In exams, check solutions against excluded denominator values.

(D=(-12)^2-4(1)(11)=100), so (x=\frac{12\pm10}{2}) gives (1) and (11). In exams, if (D) is a perfect square, simplify quickly.

The sum of roots is (-\frac{p}{8}), so (-\frac{p}{8}=-9) gives (p=72). In exams, remember the sum formula (-\frac{b}{a}).