Methods of Solving Quadratic Equations

The governing concept is that division by a variable can discard a possible zero value of that variable. First factor the equation without dividing: x² − 11x = x(x − 11), so x(x − 11) = 0. By the zero-product property, either x = 0 or x − 11 = 0, which gives x = 11. If we divide the original equation by x, we implicitly assume x ≠ 0; consequently, the solution x = 0 is lost, and only x = 11 remains in the reduced equation. Therefore option A is correct. Substitution confirms both original roots: 0² − 11(0) = 0 and 11² − 11(11) = 0. The safe method is to factor first and consider every factor separately.

(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 correct answer is option A: a=2, b=5, c=-3. Standard form means every term must be on the left and the equation must equal zero. Starting with 2x^2+5x=3, subtract 3 from both sides: 2x^2+5x-3=0. Comparing this with ax^2+bx+c=0 gives a=2, b=5, and c=-3. Option A matches all three coefficients. Option B has c=3 because it fails to change the sign when 3 is moved to the left. Option C swaps the coefficients of x^2 and x, so it does not match the equation. Option D changes the signs of the x and constant terms incorrectly. The operation must be applied to both sides, and moving a positive 3 from the right to the left makes it negative. Exam cue: first make the right side zero, then read the coefficients in order.

(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, equal real roots occur precisely when the discriminant D = b² − 4ac is zero. In x² − 10x + k = 0, the coefficients are a = 1, b = −10, and c = k. Substitution gives D = (−10)² − 4(1)(k) = 100 − 4k. Setting the discriminant equal to zero, 100 − 4k = 0, so 4k = 100 and k = 25. Thus option A is correct. Option B is merely the magnitude of the coefficient of x and does not satisfy the discriminant condition. Option C gives D = 20, not zero, while option D incorrectly uses 100 without accounting for the term 4ac. The method also confirms that the repeated root would be 5.

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

The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.

In (x^2-8x=0), (x) is a common factor and (x=0) can be a root. In exams, write (x(x-8)=0) in such cases.

Use the identity \(a^2-2ab+b^2=(a-b)^2\). Here, \(9x^2=(3x)^2\), \(25=5^2\), and the middle term is \(-30x=-2\cdot3x\cdot5\). Therefore, \(9x^2-30x+25=(3x-5)^2\), so the equation becomes \((3x-5)^2=0\). Option B would produce a positive middle term. In exams, take the square roots of the first and last terms and verify the middle term using \(\pm2ab\).

((3x-5)^2=0), so (3x-5=0) and (x=\frac{5}{3}). In exams, solve the linear equation after square form.

Bringing (8x) to the left gives (4x^2-8x-1=0). In exams, do not miss any term while making standard form.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=4\), \(b=-8\), and \(c=-1\), so \(D=(-8)^2-4(4)(-1)=64+16=80\). The value 48 results from incorrectly treating the contribution of \(-4ac\) as negative. Exam tip: when \(c\) is negative, \(-4ac\) contributes a positive value.

From (x^2-12x+20=0), (x^2-12x=-20), then adding (36) gives ((x-6)^2=16). In exams, add the same number to both sides.