Sum and product of the roots
We have seen that, in the case when a parabola crosses the \(x\)-axis, the \(x\)-coordinate of the vertex lies at the average of the intercepts. Thus, if a quadratic has two real roots \(\alpha, \beta\), then the \(x\)-coordinate of the vertex is \(\dfrac{1}{2}(\alpha+\beta)\). Now we also know that this quantity is equal to \(-\dfrac{b}{2a}\). Thus we can express the sum of the roots in terms of the coefficients \(a,b,c\) of the quadratic as \(\alpha+\beta = -\dfrac{b}{a}\).
In the case when the quadratic does not cross the \(x\)-axis, the corresponding quadratic equation \(ax^2+bx+c=0\) has no real roots, but it will have complex roots (involving the square root of negative numbers). The formula above, and other similar formulas shown below, still work in this case.
We can find simple formulas for the sum and product of the roots simply by expanding out. Thus, if \(\alpha, \beta\) are the roots of \(ax^2+bx+c=0\), then dividing by \(a\) we have
\[ x^2+\dfrac{b}{a}x+\dfrac{c}{a} = (x- \alpha)(x-\beta) = x^2-(\alpha+\beta)x+ \alpha\beta. \]
Comparing the first and last expressions we conclude that
\[ \alpha + \beta = -\dfrac{b}{a} \qquad \text{and} \qquad \alpha\beta = \dfrac{c}{a}. \]
From these formulas, we can also find the value of the sum of the squares of the roots of a quadratic without actually solving the quadratic.
Example
Suppose \(\alpha, \beta\) are the roots of \(2x^2-4x+7=0\). Find the value of \(\alpha^2+ \beta^2\) and explain why the roots of the quadratic cannot be real.
Solution
Using the formulas above, we have \(\alpha+\beta = 2\) and \(\alpha\beta = \dfrac{7}{2}\). Now \((\alpha+\beta)^2 = \alpha^2+\beta^2 + 2\alpha\beta\), so \(\alpha^2+ \beta^2 = 2^2-2\times \dfrac{7}{2}= -3\). If the roots were real, then the sum of their squares would be positive. Since the sum of their squares is \(-3\), the roots cannot be real.
Exercise 7
Find the monic quadratic with roots \(2-3\sqrt{5}\), \(2+3\sqrt{5}\).
Exercise 8
Suppose \(\alpha, \beta\) are the roots of \(2x^2-4x+7=0\).
Find the value of
- a \(\dfrac{1}{\alpha} + \dfrac{1}{\beta}\)
- b \(\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2}\).
Note that, in the previous exercise, the desired expressions are symmetric. That is, interchanging \(\alpha\) and \(\beta\) does not change the value of the expression. Such expressions are called symmetric functions of the roots.
Exercise 9
Suppose \(\alpha, \beta\) are the roots of \(ax^2+bx+c=0\). Find a formula for \((\alpha - \beta)^2\) in terms of \(a\), \(b\) and \(c\).
You may recall that the arithmetic mean of two positive numbers \(\alpha\) and \(\beta\) is \(\dfrac{1}{2}(\alpha + \beta)\), while their geometric mean is \(\sqrt{\alpha\beta}\). Thus, if a quadratic has two positive real solutions, we can express their arithmetic and geometric mean using the above formulas.
Exercise 10
Without solving the equation, find the arithmetic and geometric mean of the roots of the equation \(3x^2-17x+12=0\).
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