Mathematics » Equations and Inequalities » Solving Quadratic Equations

# Method For Solving Quadratic Equations

## Method For Solving Quadratic Equations

1. Rewrite the equation in the required form, $$a{x}^{2} + bx + c = 0$$.

2. 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 $${x}^{2} + 2x + 1 = 0$$.

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

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

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

## Example

### Question

Solve for $$x$$:

$3{x}^{2} + 2x – 1 = 0$

### 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

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

## Example

### Question

Find the roots:

$0 = -2{x}^{2} + 4x – 2$

### 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

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