93 10.8 Construct a Quadratic Equation from its Roots
It is possible to construct an equation from its roots, and the process is surprisingly simple. Consider the following:
Example 10.8.1
Construct a quadratic equation whose roots are \(x = 4\) and \(x = 6\).
This means that \(x = 4\) (or \(x - 4 = 0\)) and \(x = 6\) (or \(x - 6 = 0\)).
The quadratic equation these roots come from would have as its factored form:
\[(x - 4)(x - 6) = 0\]
All that needs to be done is to multiply these two terms together:
\[(x - 4)(x - 6) = x^2 - 10x + 24 = 0\]
This means that the original equation will be equivalent to \(x^2 - 10x + 24 = 0\).
This strategy works for even more complicated equations, such as:
Example 10.8.2
Construct a polynomial equation whose roots are \(x = \pm 2\) and \(x = 5\).
This means that \(x = 2\) (or \(x - 2 = 0\)), \(x = -2\) (or \(x + 2 = 0\)) and \(x = 5\) (or \(x - 5 = 0\)).
These solutions come from the factored polynomial that looks like:
\[(x - 2)(x + 2)(x - 5) = 0\]
Multiplying these terms together yields:
\[\begin{array}{rrrrcrrrr}
&&(x^2&-&4)(x&-&5)&=&0 \\
x^3&-&5x^2&-&4x&+&20&=&0
\end{array}\]
The original equation will be equivalent to \(x^3 - 5x^2 - 4x + 20 = 0\).
Caveat: the exact form of the original equation cannot be recreated; only the equivalent. For example, \(x^3 - 5x^2 - 4x + 20 = 0\) is the same as \(2x^3 - 10x^2 - 8x + 40 = 0\), \(3x^3 - 15x^2 - 12x + 60 = 0\), \(4x^3 - 20x^2 - 16x + 80 = 0\), \(5x^3 - 25x^2 - 20x + 100 = 0\), and so on. There simply is not enough information given to recreate the exact original—only an equation that is equivalent.
Questions
Construct a quadratic equation from its solution(s).
- 2, 5
- 3, 6
- 20, 2
- 13, 1
- 4, 4
- 0, 9
- \(\dfrac{3}{4}, \dfrac{1}{4}\)
- \(\dfrac{5}{8}, \dfrac{5}{7}\)
- \(\dfrac{1}{2}, \dfrac{1}{3}\)
- \(\dfrac{1}{2}, \dfrac{2}{3}\)
- ± 5
- ± 1
- \(\pm \dfrac{1}{5}\)
- \(\pm \sqrt{7}\)
- \(\pm \sqrt{11}\)
- \(\pm 2\sqrt{3}\)
- 3, 5, 8
- −4, 0, 4
- −9, −6, −2
- ± 1, 5
- ± 2, ± 5
- \(\pm 2\sqrt{3}, \pm \sqrt{5}\)
