Methods of Solving Quadratic Equations

(14x^2-25x+6=(7x-3)(2x-2)), so the roots are (\frac{3}{7}) and (1). In exams, also check by removing any common factor if present.

Here (ac=420) and (-28+(-15)=-43), so the correct split is (-28x-15x). In exams, even when (ac) is large, match both sum and product.

(20x^2-43x+21=(5x-7)(4x-3)), so the roots are (\frac{7}{5}) and (\frac{3}{4}). In exams, do not invert fractional roots.

For equal roots, (D=0), so (4(k+3)^2-4(k^2-16)=0) must be expanded carefully; a wrong expansion changes the answer. In exams, recheck parameter expansion.

(D=4(k+3)^2-4(k^2-16)=0) gives ((k+3)^2=k^2-16), so (6k+25=0) and (k=-\frac{25}{6}). In exams, handle the constant term carefully after expansion.

The sum of roots is (4), and (\frac{p+4}{5}=4), so (p=16). In exams, use (-\frac{b}{a}) for the sum.

First \(x^2-\frac{10}{3}x+\frac{8}{9}=0\) is obtained, then \(\left(x-\frac{5}{3}\right)^2=\frac{17}{9}\). In exams, divide by (a) first when \(a\neq1\).

Since \(\left(x-\frac{5}{3}\right)^2=\frac{17}{9}\), \(x=\frac{5\pm\sqrt{17}}{3}\). In exams, write the square root with the denominator correctly.

For real and equal roots, (D=0), so (400-4m=0) gives (m=100). In exams, equal roots indicate (D=0).

The first equation has roots (\frac{3}{2},\frac{5}{3}), and the second has roots (\frac{3}{2},\frac{6}{5}). In exams, solve both equations separately for the common root.

The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.

The other root is (13), so (q=8\times13=104). In exams, when (a=1), the constant term is the product of roots.

(6x^2+x-2=(3x+2)(2x-1)), so (x=\frac{1}{2},-\frac{2}{3}) is correct. In exams, change signs carefully from factors.

Since (11=(\sqrt{11})^2) and the middle term is (-2\sqrt{11}x), it is ((x-\sqrt{11})^2). In exams, identify perfect squares even with irrational coefficients.

Direct answer: option A, x=√11, is the repeated root. Start with x²−2√11x+11=0. Since (x−√11)²=x²−2√11x+11, the equation becomes (x−√11)²=0. A square is zero only when its base is zero, so x−√11=0 and x=√11. The root occurs twice, but the value is one repeated root. B, x=−√11, has the wrong sign; substituting it would not make the factor x−√11 zero. C, x=11, is not the square-root value and does not satisfy the equation. D, x=−11, also has the wrong sign and magnitude. Equivalently, for ax²+bx+c, the repeated-root form is (x−r)², whose middle coefficient is −2r; here r=√11. Memory cue: recognize the perfect square (x−√11)².

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

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

Dividing the whole equation by (25) gives (x^2-(a+b)x+ab=0). In exams, removing the common factor first shortens the solution.

Since (x^4=(x^2)^2=y^2), the new equation is (y^2-17y+16=0). In exams, use substitution to form a quadratic.

From (y^2-17y+16=0), (y=1,16), so (x^2=1,16) and (x=\pm1,\pm4). In exams, do not forget to return to (x).

Multiplying both sides by (5x) gives (5+5x^2=26x), that is (5x^2-26x+5=0). In exams, remember the condition (x\neq0).

(5x^2-26x+5=(5x-1)(x-5)), so (x=\frac{1}{5}) and (5). In exams, check whether obtained roots are valid in the original equation.

Cross multiplication gives ((x+4)^2=25x), so (x^2+8x+16-25x=0), and (x^2-17x+16=0). In exams, cross multiply carefully.

(x^2-17x+16=(x-1)(x-16)), so (x=1) and (x=16). In exams, check solutions against excluded denominator values.

(D=(-10)^2-4(1)(7)=72), so (x=\frac{10\pm6\sqrt{2}}{2}=5\pm3\sqrt{2}). In exams, simplify the square root.