Answer:
[tex]\displaystyle x=\frac{5}{2}, x=\frac{2}{5}[/tex]
Step-by-step explanation:
Quadratic Equations
The quadratic equation has the following general form
[tex]ax^2+bx+c=0[/tex]
It can be solved by several methods, including the well-known quadratic formula
[tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
The question requires us to find a number that added with its reciprocal results 2 9/10. If x is the required number
[tex]\displaystyle x+\frac{1}{x}=2\ \frac{9}{10}=\frac{29}{10}[/tex]
Operating
[tex]\displaystyle \frac{x^2+1}{x}=\frac{29}{10}[/tex]
Rearranging
[tex]10(x^2+1)=29x[/tex]
[tex]10x^2-29x+10=0[/tex]
Comparing with the general quadratic equation, we have
[tex]a=10,\ b=-29,\ c=10[/tex]
Using the formula
[tex]\displaystyle x=\frac{29\pm \sqrt{(-29)^2-4(10)(10)}}{2(10)}[/tex]
[tex]\displaystyle x=\frac{29\pm \sqrt{441}}{20}=\frac{29\pm 21}{20}[/tex]
It gives us two possible solutions
[tex]\displaystyle \boxed{x=\frac{5}{2}, x=\frac{2}{5}}[/tex]
Both are valid solutions since
[tex]\displaystyle \frac{5}{2}+\frac{2}{5}=\frac{29}{10}[/tex]
