Question

solving simple quadratic equations practice
complete this assessment to review what youve learned. it will not count toward your grade
solve ( x^2 = \frac{16}{25} ) by inspection. there are two real solutions. enter the lesser number first. leave the answers in simplest fraction form. (1 point)

check answer remaining attempts: 3

Explanation:

Step1: Recall the square root property

If \( x^2 = a \) (where \( a\geq0 \)), then \( x=\pm\sqrt{a} \).

Step2: Apply the property to the given equation

Given \( x^2=\frac{16}{25} \), we take the square root of both sides. So \( x = \pm\sqrt{\frac{16}{25}} \).

Step3: Simplify the square root

We know that \( \sqrt{\frac{16}{25}}=\frac{\sqrt{16}}{\sqrt{25}} \) (by the property \( \sqrt{\frac{m}{n}}=\frac{\sqrt{m}}{\sqrt{n}} \) for \( m\geq0,n > 0 \)). Since \( \sqrt{16} = 4 \) and \( \sqrt{25}=5 \), we have \( \sqrt{\frac{16}{25}}=\frac{4}{5} \). So \( x=\pm\frac{4}{5} \), which means \( x = \frac{4}{5} \) or \( x=-\frac{4}{5} \).

Step4: Identify the lesser solution

Comparing \( -\frac{4}{5} \) and \( \frac{4}{5} \), we know that \( -\frac{4}{5}<\frac{4}{5} \).

Answer:

\(-\frac{4}{5}, \frac{4}{5}\)