Methods of Solving Quadratic Equations

\((x+5)^2=x^2+2(5)x+5^2=x^2+10x+25\). Comparing this with \(x^2+10x+k\) gives \(k=25\). The distractor 5 is the constant inside the bracket, whereas \(k\) is its square. Exam tip: in \((x+a)^2=x^2+2ax+a^2\), the constant term is \(a^2\).

In completing square method, the quadratic part is made into the form ((x+p)^2). In exams, this method is useful when simple factors are not found quickly.

(25x^2-1=(5x)^2-1^2), so ((5x-1)(5x+1)=0) is correct. In exams, quickly identify the difference of squares.

(2x^2-7x+3=(2x-1)(x-3)), so (x=3) and (x=\frac{1}{2}). In exams, solve each linear factor separately.

Here (ac=18) and (9+2=11), so (11x) is split as (9x+2x). In exams, check both sum (b) and product (ac).

(3x^2-2x-1=(3x+1)(x-1)), so the roots are (1) and (-\frac{1}{3}). In exams, form mixed-sign factors carefully.

(-5+3=-2) and (-5\times3=-15), so ((x-5)(x+3)) is correct. In exams, match both product and sum.

Here (D=(-4)^2-4(1)(-5)=36), so (x=\frac{4\pm6}{2}). In exams, do not forget the negative sign of (c) while using the formula.

Completing the square means transforming the quadratic expression into a perfect-square expression without changing the equation. Start with x² + 6x + 1 = 0 and move the constant term: x² + 6x = −1. Half of the coefficient of x is 6/2 = 3, and its square is 9. Add 9 to both sides: x² + 6x + 9 = −1 + 9 = 8. The left side becomes (x + 3)², so (x + 3)² = 8. Therefore option A is correct. Option B uses the full coefficient instead of half, option C has an incorrect right side, and option D has the wrong sign inside the square.

Rewrite the equation as \(x^2+6x+9=8\), so \((x+3)^2=8\). Thus, \(x+3=\pm\sqrt{8}=\pm2\sqrt{2}\), giving \(x=-3\pm2\sqrt{2}\). Option D misses the factor 2 while simplifying \(\sqrt{8}\). Exam tip: to complete the square, add the square of half the coefficient of \(x\).

The standard form is (2x^2+5x-3=0), so (a=2,b=5,c=-3). In exams, bring all terms to one side first.

(2x^2+5x-3=(2x-1)(x+3)), so (x=\frac{1}{2}) and (-3). In exams, write fractional roots correctly.

Using the identity \((a-b)^2=a^2-2ab+b^2\), we get \(4x^2-12x+9=(2x-3)^2\). Hence the equation becomes \((2x-3)^2=0\). In option B, the middle term would be positive, whereas the given middle term is \(-12x\). In exams, take the square roots of the first and last terms and then verify the middle term.

((2x-3)^2=0), so (2x-3=0) and (x=\frac{3}{2}). In exams, write the repeated root with the correct value.

The correct form is (5x(x-4)=0), giving (x=0) and (x=4). In exams, dividing directly by the variable can miss (x=0).

(3x^2+5x-2=(3x-1)(x+2)), so the roots are (\frac{1}{3}) and (-2). In exams, solve (3x-1=0) carefully.

In the standard form \(ax^2+bx+c=0\), we have \(a=2\), \(b=-4\), and \(c=-3\). Thus, \(D=b^2-4ac=(-4)^2-4(2)(-3)=16+24=40\). Since \(c\) is negative, the term \(-4ac\) becomes positive. Exam tip: identify the signs of \(a\), \(b\), and \(c\) before substituting; taking only \(b^2=16\) is incorrect.

The formula gives (x=\frac{4\pm\sqrt{40}}{4}=1\pm\frac{\sqrt{10}}{2}). In exams, simplify the final form.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. The equation has equal real roots exactly when its discriminant is zero. Here a = 1, b = −10, and c = k. Therefore D = (−10)² − 4(1)(k) = 100 − 4k. Set this equal to zero: 100 − 4k = 0, so 4k = 100 and k = 25. Hence option A is correct. Option B is merely the magnitude of the x coefficient, option C does not make the discriminant zero, and option D results from using 10² without the factor 4ac.

For equal roots, (k^2-64=0), so (k=\pm8), and (k>0) gives (k=8). In exams, apply the given condition.