Quadratic equation using completing square method

Examples

As an example, let's solve the quadratic equation: $2x² + 5x - 3 = 0$

Explanation:

1. Write the equation in the form of $ax² + bx + c = 0$: $2x² + 5x - 3 = 0$

2. Divide the entire equation by 'a' i.e. $2$: $\frac22x² + \frac{5}{2}x - \frac32 = 0 \\ \\x² + \frac{5}{2}x - \frac32 = 0$

3. Move the constant term $-\frac32$ to the right-hand side: $x² + \frac52x = \frac32$

4. Take half of the coefficient of $x$ term and square it: $\frac52 \times \frac12= \frac54$

$(\frac54)^2$ 5. Complete the square by adding and subtracting $(\frac54)^²$ on the left-hand side: $x² + \frac52x + (\frac54)^2 - (\frac54)^2 = \frac32$ 6. Simplify the left-hand side of the equation:

$(x + \frac54)^2 - \frac{25}{16} = \frac32$

$(x + \frac54)^2 = \frac32+ \frac{25}{16}$

$(x + \frac54)^2 = \frac{49}{16}$ 7. Take the square root of both sides of the equation: $\sqrt{(x + \frac54)^2} = \pm \sqrt{\frac{49}{16}}$ On right side, square and root will be cancelled, $(x + \frac54) = \pm \sqrt{\frac{49}{16}}$ 8. Solve for $x$:

$(x + \frac54) = \pm \frac{7}{4}$

$x = - \frac54\pm \frac{7}{4}$

$x = - \frac54 + \frac{7}{4}$ ; $x = - \frac54- \frac{7}{4}$

$x = \frac{2}{4}$ ; $x = - \frac{12}{4}$

$x = \frac{1}{2}$ ; $x = -3$

Therefore, the solutions of the given quadratic equation are $x = -3$ and $x = \frac{1}{2}$

Solution:

$2x² + 5x - 3 = 0$

$\frac22x² + \frac{5}{2}x - \frac32 = 0 \\ \\x² + \frac{5}{2}x - \frac32 = 0$

$x² + \frac52x = \frac32$

$(\frac52 \times \frac12)^2=( \frac54)^2$

$x² + \frac52x + (\frac54)^2 - (\frac54)^2 = \frac32$

$(x + \frac54)^2 - \frac{25}{16} = \frac32$

$(x + \frac54)^2 = \frac32+ \frac{25}{16}$

$(x + \frac54)^2 = \frac{49}{16}$ taking square root on both sides $(x + \frac54) = \pm \sqrt{\frac{49}{16}}$ solving for $x$ $(x + \frac54) = \pm \frac{7}{4}$

$x = - \frac54\pm \frac{7}{4}$

$x = - \frac54 + \frac{7}{4}$ ; $x = - \frac54- \frac{7}{4}$

$x = \frac{2}{4}$ ; $x = - \frac{12}{4}$

$x = \frac{1}{2}$ ; $x = -3$