The kuadarat equation has a general form $ ax^2 + bx + c = 0 $ with $ a \neq 0 $. The solution to the quadratic equation can be done in the following 5 ways. But, what is commonly used is factoring and the ABC formula. ABC formula is used when the method of factoring cannot be done easily. Consider one of the following examples that can be solved in the following 5 ways.

1. Solving quadratic equations by factori;

2. Solving quadratic equations by complete perfect squares;

3. Solving quadratic equations using the quadratic formula;

4. Solving quadratic equations by Substitution; and

5. Solving quadratic equations by difference of 2 squares

Determine the solution equation $ x^2 + 4x - 5 = 0 $

Method 1:

$ \begin{align} x^2 + 4x - 5 & = 0 \\ (x - 1) (x + 5) & = 0 \\ x - 1 = 0 \ or \ & x + 5 = 0 \\ x = 1 \ or \ & x = -5 \end{align} $

So, the set of resolutions is {1, -5}.

Method 2:

$ \begin{align} x^2 + 4x -5 & = 0 \\ x^2 + 4x & = 5 \\ x^2 + 4x + (\frac{b}{2})^2 & = 5+ (\frac{b}{2})^2 \\ x^2 + 4x + (\frac{4}{2})^2 & = 5+ (\frac{4}{2})^2 \\ x^2 + 4x + 2^2 & = 5+ 2^2 \\ (x +2)^2 & = 9 \\ x +2 & = \pm \sqrt{9} \\ x +2 & = \pm 3 \\ x & = -2 \pm 3 \\ x_1 & = -2 +3 = 1 \\ x_2 & = –2 - 3 = -5 \end{align} $

So, the set of resolutions is {1, -5}

Method 3:

$ \Rightarrow \\ \begin{align} x_{1; 2} & = \frac{-(4) \pm \sqrt{(4)^2-(4)(1)(- 5)}}{2(1)} \\ & = \frac{-4 \pm \sqrt{16 + 20}}{2} \\ & = \frac{-4 \pm \sqrt{36}}{2} \\ & = \frac{-4 \pm 6}{2} \\ \\ x_1 & = \frac{-4 + 6}{2} \\ & = \frac{2 }{2} \\ & = 1 \\ \\ x_2 & = \frac{-4-6}{2} \\ & = \frac{-10}{2} \\ & = - 5 \end{align} $

So, the set of resolutions is {1, -5}

Method 4:

Substitution:

$ \begin{align} x & = y- \frac{b}{2a} \\ & = y- \frac{4}{2 \dot 1} \\ & = y-2 \end{align} $

to $ x ^ 2 + 4x - 5 = 0 $

obtained:

$ \begin{align} (y-2)^2 +4 (y-2) - 5 & = 0 \\ y^2-4y + 4 + 4y-8-5 & = 0 \\ y^2-9 & = 0 \\ y^2 & = 9 \\ y & = \pm 3 \end{align} $

so,

$ \begin{align} x_1 & = 3-2 \\ & = 1 \\ x_2 & = –3 - 2 \\ & = –5 \end{align} $

So, the set of resolutions is {1, -5}.

Method 5:

We can change each quadratic equation to the difference of 2 squares as follows.

$ \begin{align} (x + p)^2 -q^2 & = 0 \\ x^2 + 2px + p^2 - q^2 & = 0 \\ x^2 + 4x –5 & = 0 \end{align} $

Obtained:

$ \begin{align} 2p & = 4 \\ p & = 2 \end{align} $

$ \begin{align} p^2-q^2 & = –5 \\ 4 - q^2 & = –5 \\ q^2 & = 9 \Rightarrow q = 3 \end{align} $

So that,

$ \begin{align} (x + p)^2 - q^2 & = 0 \\ (x + p + q) (x + p - q) & = 0 \\ (x + 2 + 3) (x + 2-3) & = 0 \\ (x + 5) (x -1) & = 0 \\ x = –5 \ or \ x & = 1 \end{align} $ ,

So, the set of resolutions is {1, -5}.

1. Solving quadratic equations by factori;

2. Solving quadratic equations by complete perfect squares;

3. Solving quadratic equations using the quadratic formula;

4. Solving quadratic equations by Substitution; and

5. Solving quadratic equations by difference of 2 squares

__Question example__:Determine the solution equation $ x^2 + 4x - 5 = 0 $

Method 1:

**Solving quadratic equations by factori**$ \begin{align} x^2 + 4x - 5 & = 0 \\ (x - 1) (x + 5) & = 0 \\ x - 1 = 0 \ or \ & x + 5 = 0 \\ x = 1 \ or \ & x = -5 \end{align} $

So, the set of resolutions is {1, -5}.

Method 2:

**Solving quadratic equations by complete perfect squares**$ \begin{align} x^2 + 4x -5 & = 0 \\ x^2 + 4x & = 5 \\ x^2 + 4x + (\frac{b}{2})^2 & = 5+ (\frac{b}{2})^2 \\ x^2 + 4x + (\frac{4}{2})^2 & = 5+ (\frac{4}{2})^2 \\ x^2 + 4x + 2^2 & = 5+ 2^2 \\ (x +2)^2 & = 9 \\ x +2 & = \pm \sqrt{9} \\ x +2 & = \pm 3 \\ x & = -2 \pm 3 \\ x_1 & = -2 +3 = 1 \\ x_2 & = –2 - 3 = -5 \end{align} $

So, the set of resolutions is {1, -5}

Method 3:

**Solving quadratic equations using the quadratic formula**Formula: $ x_{1; 2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$ x^2 + 4x - 5 = 0; \ \ \ a = 1, \ b = 4, \ c = -5 $

$ \Rightarrow \\ \begin{align} x_{1; 2} & = \frac{-(4) \pm \sqrt{(4)^2-(4)(1)(- 5)}}{2(1)} \\ & = \frac{-4 \pm \sqrt{16 + 20}}{2} \\ & = \frac{-4 \pm \sqrt{36}}{2} \\ & = \frac{-4 \pm 6}{2} \\ \\ x_1 & = \frac{-4 + 6}{2} \\ & = \frac{2 }{2} \\ & = 1 \\ \\ x_2 & = \frac{-4-6}{2} \\ & = \frac{-10}{2} \\ & = - 5 \end{align} $

So, the set of resolutions is {1, -5}

Method 4:

**Solving quadratic equations by Substitution**$ x = y- \frac{b}{2a} $Substitution:

$ \begin{align} x & = y- \frac{b}{2a} \\ & = y- \frac{4}{2 \dot 1} \\ & = y-2 \end{align} $

to $ x ^ 2 + 4x - 5 = 0 $

obtained:

$ \begin{align} (y-2)^2 +4 (y-2) - 5 & = 0 \\ y^2-4y + 4 + 4y-8-5 & = 0 \\ y^2-9 & = 0 \\ y^2 & = 9 \\ y & = \pm 3 \end{align} $

so,

$ \begin{align} x_1 & = 3-2 \\ & = 1 \\ x_2 & = –3 - 2 \\ & = –5 \end{align} $

So, the set of resolutions is {1, -5}.

Method 5:

**Solving quadratic equations by Difference 2 squares**We can change each quadratic equation to the difference of 2 squares as follows.

$ \begin{align} (x + p)^2 -q^2 & = 0 \\ x^2 + 2px + p^2 - q^2 & = 0 \\ x^2 + 4x –5 & = 0 \end{align} $

Obtained:

$ \begin{align} 2p & = 4 \\ p & = 2 \end{align} $

$ \begin{align} p^2-q^2 & = –5 \\ 4 - q^2 & = –5 \\ q^2 & = 9 \Rightarrow q = 3 \end{align} $

So that,

$ \begin{align} (x + p)^2 - q^2 & = 0 \\ (x + p + q) (x + p - q) & = 0 \\ (x + 2 + 3) (x + 2-3) & = 0 \\ (x + 5) (x -1) & = 0 \\ x = –5 \ or \ x & = 1 \end{align} $ ,

So, the set of resolutions is {1, -5}.

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