Chapter 8.6, Problem 36PPS

To determine

Tosolve the given equation and check the solution.

Answer to Problem 36PPS

  $a=-\frac{5}{7},-\frac{5}{7}$

Explanation of Solution

Given:

The given equation is:

  $a^2+\frac{10}{7}a+\frac{25}{49}=0$

Calculation:

The equation can be solved as:

  $\begin{array}{l}{a}^2+\frac{10}{7}a+\frac{25}{49}=0\\ a^2+2\cdot a\cdot \frac{5}{7}+{\left(\frac{5}{7}\right)}^2=0\\ {\left(a+\frac{5}{7}\right)}^2=0\\ a=-\frac{5}{7},-\frac{5}{7}\end{array}$

Check:

Put $a=-\frac{5}{7},-\frac{5}{7}$ ,

  $\begin{array}{c}{a}^2+\frac{10}{7}a+\frac{25}{49}=0\\ {\left(-\frac{5}{7}\right)}^2+\frac{10}{7}\left(-\frac{5}{7}\right)+\frac{25}{49}=0\\ \frac{25}{49}-\frac{50}{49}+\frac{25}{49}=0\\ \frac{50}{49}-\frac{50}{49}=0\\ 0=0\end{array}$

Hence, the solution is $a=-\frac{5}{7},-\frac{5}{7}$ .