The solutions to any quadratic equation can be found by substituting the values a, b, c into the quadratic formula.

The solutions to any quadratic equation $$ax^{2}+bx+c=0$$ can be found by substituting the values $$a,b,c$$ into the quadratic formula:

$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$ Solutions may be real or complex numbers.

In this module we only consider real solutions.

### Example 1

Solve the equation: $x^{2}-5x+4=0.$

Solution:

We have1 Note that $$x^{2}=1\times x^{2}$$ and so $$a=1$$ . $$a=1,\ \,b=-5$$ and $$c=4.$$ Substituting into the formula we get: \begin{align*} x & =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\ & =\frac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(4)}}{2(1)}\\ & =\frac{5\pm\sqrt{25-16}}{2}\\ & =\frac{5\pm\sqrt{9}}{2}\\ & =\frac{5\pm3}{2}\\ & =\frac{5+3}{2}\ \textrm{ or }\ \frac{5-3}{2}\\ & =\frac{8}{2}\ \textrm{ or }\ \frac{2}{2}\\ & =4\ \textrm{ or }\ 1. \end{align*} The solution is $$x=1$$  or  $$x=4$$.

### Example 2

Solve the equation: $x^{2}+x+10=0.$

Solution:

We have $$a=1,\,b=1$$ and $$c=10.$$ Substituting into the formula we get: \begin{align*} x & =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\ & =\frac{-1\pm\sqrt{1^{2}-4(1)(10)}}{2(1)}\\ & =\frac{-1\pm\sqrt{1-40}}{2}\\ & =\frac{-1\pm\sqrt{-39}}{2}. \end{align*} Since the $$\sqrt{-39}$$ does not exist the equation has no solution.2 There is a solution if we use complex numbers but we are looking for real solutions.

### Discriminant

The part of the quadratic formula which is under the radical sign ($$b^{2}-4ac)$$ is called the discriminant. Its value determines the number of solutions and whether they will be rational or irrational.

• If $$b^{2}-4ac<0$$ there are no (real) solutions. This means the graph of $$y=ax^{2}+bx+c$$ doesn’t touch or cut the $$x$$-axis.
• If $$b^{2}-4ac=0$$ there is one solution. This means the graph of $$y=ax^{2}+bx+c$$ touches the $$x$$-axis at this value of $$x$$.
• If $$b^{2}-4ac>0$$ there are two solutions. This means the graph of $$y=ax^{2}+bx+c$$ cuts the $$x$$-axis at these values of $$x$$.
• If $$b^{2}-4ac\quad$$is a perfect square the solutions will be rational .

3 A number is a perfect square if its square root is an integer. Integers are whole numbers: $$\ldots,-2,-1,0,1,2,\ldots$$
4 A rational number can be expressed as a fraction $$p/q$$ where $$p$$ and $$q\neq0$$ are integers.

### Exercise

Solve the following quadratic equations for real solutions (if possible):

$\begin{array}{llllllll} \textrm{1.} & x-4x-7=0 & & \textrm{} & & \textrm{} & \textrm{2.} & x^{2}-2x-2=0\\ \\ 3. & x^{2}+6x-9=0 & & & & & \textrm{4.} & x^{2}-x-7=0\\ \\ 5. & 4x^{2}-12x+6=0 & & & & & \textrm{6.} & 2x^{2}+x-2=0\\ \\ \textrm{7.} & x^{2}-4x+2=0 & & & & & \textrm{8.} & 2x^{2}-3x+2=0\\ \\ \textrm{9.} & 3x^{2}+5x-7=0 & & & & & \textrm{10.} & 3x^{2}=x+1\\ \\ \textrm{11.} & 4x^{2}+x+3=0 & & & & & \textrm{12.} & \frac{3x+1}{2}=\frac{x+1}{x} \end{array}$

$\begin{array}{llllllll} \textrm{1.} & x=\frac{4\pm\sqrt{44}}{2}=2\pm\sqrt{11} & & \textrm{} & & \textrm{} & \textrm{2.} & x=\frac{2\pm\sqrt{12}}{2}=1\pm\sqrt{3}\\ \\ \textrm{3.} & x=\frac{-6\pm\sqrt{72}}{2}=-3\pm3\sqrt{2} & & & & & 4. & x=\frac{1\pm\sqrt{29}}{2}\\ \\ 5. & x=\frac{12\pm\sqrt{48}}{8}=\frac{3\pm\sqrt{3}}{2} & & & & & \textrm{6.} & x=\frac{-1\pm\sqrt{17}}{4}\\ \\ \textrm{7.} & x=\frac{4\pm\sqrt{8}}{2}=2\pm\sqrt{2} & & & & & \textrm{8.} & no\:solution\\ \\ \textrm{9.} & x=\frac{-5\pm\sqrt{109}}{6} & & & & & \textrm{10.} & x=\frac{1\pm\sqrt{13}}{6}\\ \\ \textrm{11.} & no\:solution & & & & & \textrm{12.} & x=-\frac{2}{3},\quad x=1 \end{array}$