## Method For Solving Quadratic Equations

Rewrite the equation in the required form, \(a{x}^{2} + bx + c = 0\).

Divide the entire equation by any common factor of the coefficients to obtain an equation of the form \(a{x}^{2} + bx + c = 0\), where \(a\), \(b\) and \(c\) have no common factors. For example \(2{x}^{2} + 4x + 2 = 0\) can be written as \({x}^{2} + 2x + 1 = 0\).

Factorise \(a{x}^{2} + bx + c = 0\) to be of the form \((rx + s)(ux + v) = 0\).

The two solutions are \((rx + s) = 0\) or \((ux + v) = 0\), so \(x= -\dfrac{s}{r}\) or \(x = -\dfrac{v}{u}\), respectively.

Check the answer by substituting it back into the original equation.

## Example

### Question

Solve for \(x\):

\[3{x}^{2} + 2x – 1 = 0\]

### The equation is already in the required form, \(a{x}^{2} + bx + c = 0\)

### Factorise

\[(x + 1)(3x – 1) = 0\]

### Solve for both factors

We have

\begin{align*} x + 1 & = 0 \\ \therefore x & = -1 \end{align*}

OR

\begin{align*} 3x – 1 & = 0 \\ \therefore x & = \cfrac{1}{3} \end{align*}

### Check both answers by substituting back into the original equation

### Write the final answer

The solution to \(3{x}^{2} + 2x – 1 = 0\) is \(x = -1\) or \(x = \cfrac{1}{3}\).

## Example

### Question

Find the roots:

\[0 = -2{x}^{2} + 4x – 2\]

### The equation is already in the required form, \(a{x}^{2} + bx + c = 0\)

### Divide the equation by common factor \(-\text{2}\)

\begin{align*} -2{x}^{2} + 4x – 2 & = 0 \\ {x}^{2} – 2x + 1 & = 0 \end{align*}

### Factorise

\begin{align*} (x – 1)(x – 1) & = 0 \\ {(x – 1)}^{2} & = 0 \end{align*}

### The quadratic is a perfect square

This is an example of a special situation in which there is only one solution to the quadratic equation because both factors are the same.

\begin{align*} x – 1 & = 0 \\ \therefore x & = 1 \end{align*}

### Check the answer by substituting back into the original equation

### Write final answer

The solution to \(0 = -2{x}^{2} + 4x – 2\) is \(x = 1\).