The discriminant
Roots can occur in a parabola in 3 different ways as shown in the diagram below:
In diagram A, we can see that this parabola has 2 roots, diagram B has 1 root and diagram C has no roots.
What type of roots the equation has can be shown by the discriminant.
The discriminant for a quadratic equation \(a{x^2} + bx + c = 0\) is \({b^2} - 4ac\). And the types of root the equation has can be worked out as follows:
- If \({b^2} - 4ac\textgreater0\), the roots are real and unequal (diagram A)
- If \({b^2} - 4ac = 0\), the roots are real and equal (diagram B)
- If \({b^2} - 4ac\textless0\), the roots are non-real (diagram C)
Example
Find the nature of the roots of \(2{x^2} + 3x - 6 = 0\)
Solution
\(a = 2,\,b = 3,c = - 6\)
\({b^2} - 4ac\)
\(= {(3)^2} - 4(2)( - 6)\)
\(= 9 - ( - 48)\)
\(= 57\)
Since \({b^2} - 4ac\textgreater0\), the roots are real and unequal.
Question
For the quadratic function \(y = (2x + 3)(x - 5)\) determine the nature of the roots and then solve.
Firstly we need to remove the brackets using FOIL:
- Multiply the first term in each set of brackets
- Multiply the outer term in each set of brackets
- Multiply the inner term in each set of brackets
- Multiply the last term in each set of brackets
\(y = (2x + 3)(x - 5)\)
\(y = 2{x^2} - 10x + 3x - 15\)
\(y = 2{x^2} - 7x - 15\)
Using the discriminant to find the nature:
\(a = 2,\,b = - 7,c = - 15\)
\({b^2} - 4ac\)
\(= {( - 7)^2} - (4 \times 2 \times - 15)\)
\(= 49 - ( - 120)\)
\(= 49 + 120\)
\(= 169\textgreater0\) therefore roots are real and unequal.
To solve the quadratic equation, we make \(y = 0\) and since the question already gave us the quadratic in its factorised form, we will use this to solve to find values for \(x\).
\((2x + 3)(x - 5) = 0\)
\(2x + 3 = 0\)
\(2x = - 3\)
\(x = \frac{{ - 3}}{2}\)
OR:
\(x - 5 = 0\)
\(x = 5\)
Question
Find the value(s) of \(k\) if \({x^2} + 2kx + 36 = 0\) has one real root.
If the quadratic has one real root, then \({b^2} - 4ac = 0\).
Therefore we have to solve this using \(a = 1,\,b = 2k\,and\,c = 36\)
\({b^2} - 4ac = 0\)
\({(2k)^2} - (4 \times 1 \times 36) = 0\)
\(4{k^2} - 144 = 0\)
\(4{k^2} = 144\)
\({k^2} = \frac{{144}}{4}\)
\({k^2} = 36\)
\(k = \sqrt {36}\)
\(k = \pm 6\)
