To construct a quadratic equation given its roots, we need to use the fact that for a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], if [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots, then:
1. The sum of the roots, [tex]\(\alpha + \beta = -\frac{b}{a}\)[/tex]
2. The product of the roots, [tex]\(\alpha \beta = \frac{c}{a}\)[/tex]
Given the roots [tex]\(\frac{4}{5}\)[/tex] and [tex]\(-2 \frac{2}{3}\)[/tex], we first convert these numbers to a common fraction form:
- [tex]\(\frac{4}{5}\)[/tex] is already in fraction form.
- [tex]\(-2 \frac{2}{3}\)[/tex] can be converted to an improper fraction:
[tex]\[
-2 \frac{2}{3} = -\left(2 + \frac{2}{3}\right) = -\left(\frac{6}{3} + \frac{2}{3}\right) = -\frac{8}{3}
\][/tex]
Now, let's denote the roots as:
[tex]\[
\alpha = \frac{4}{5}, \quad \beta = -\frac{8}{3}
\][/tex]
Step 1: Calculate the sum of the roots.
Sum of the roots [tex]\(\alpha + \beta\)[/tex] is:
[tex]\[
\alpha + \beta = \frac{4}{5} + \left( -\frac{8}{3} \right)
\][/tex]
To add these fractions, we need a common denominator. The least common multiple of 5 and 3 is 15. Converting both fractions to have this common denominator:
[tex]\[
\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}, \quad -\frac{8}{3} = \frac{-8 \times 5}{3 \times 5} = \frac{-40}{15}
\][/tex]
Now we can add them:
[tex]\[
\frac{12}{15} + \frac{-40}{15} = \frac{12 - 40}{15} = \frac{-28}{15}
\][/tex]
Step 2: Calculate the product of the roots.
Product of the roots [tex]\(\alpha \beta\)[/tex] is:
[tex]\[
\alpha \beta = \left(\frac{4}{5}\right) \left(-\frac{8}{3}\right)
\][/tex]
Multiplying the fractions:
[tex]\[
\left(\frac{4}{5}\right) \left(-\frac{8}{3}\right) = \frac{4 \times (-8)}{5 \times 3} = \frac{-32}{15}
\][/tex]
Step 3: Write the quadratic equation using the sum and product of the roots.
In general, the quadratic equation based on the sum and product of the roots is given by:
[tex]\[
x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0
\][/tex]
Plugging in our values:
[tex]\[
x^2 - \left(\frac{-28}{15}\right)x + \left(\frac{-32}{15}\right) = 0
\][/tex]
Simplifying the signs:
[tex]\[
x^2 + \frac{28}{15}x - \frac{32}{15} = 0
\][/tex]
Thus, the quadratic equation whose roots are [tex]\(\frac{4}{5}\)[/tex] and [tex]\(-2\frac{2}{3}\)[/tex] is:
[tex]\[
x^2 + \frac{28}{15}x - \frac{32}{15} = 0
\][/tex]
