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Which statement(s) true when solving the equation for x ? CHOOSE TWO.

x^2 =81/144


(2 points)
Responses

You would use the inverse operation of dividing both sides by 2
You would use the inverse operation of dividing both sides by 2

You would use the inverse operation of square root
You would use the inverse operation of square root

You would use the inverse operation of multiplying both sides by itself
You would use the inverse operation of multiplying both sides by itself

You would use the inverse operation of dividing both sides by 4
You would use the inverse operation of dividing both sides by 4

There is no solution for x in this equation.
There is no solution for x in this equation.

there would be one solution for x;

x = 81/72 = 9/8

there would be one solution for x; x = 81/72 = 9/8

there would be two solutions for x;

x = 9/12 = +3/4 and x = - 9/12 = - 3/4

there would be two solutions for x; x = 9/12 = +3/4 and x = - 9/12 = - 3/4

there would be one solution for x;

x = 9/12 = 3/4

there would be one solution for x; x = 9/12 = 3/4

there would be two solutions for x;

x = 81/72 = 9/8 and x = - 81/72 = -9/8

1 answer

To solve the equation \( x^2 = \frac{81}{144} \), we should first simplify \(\frac{81}{144}\):

\[ \frac{81}{144} = \frac{9}{16} \]

This leads us to the equation:

\[ x^2 = \frac{9}{16} \]

Next, to solve for \(x\), we use the square root operation:

\[ x = \pm \sqrt{\frac{9}{16}} = \pm \frac{3}{4} \]

Now we can analyze the provided responses to identify the two true statements:

  1. You would use the inverse operation of square root.

    • This is true since we take the square root of both sides to solve for \(x\).

  2. There would be two solutions for (x; x = \frac{9}{12} = \frac{3}{4} \text{ and } x = -\frac{9}{12} = -\frac{3}{4}.

    • This is also true because when solving for \(x\) we find two values: \( \frac{3}{4} \) and \(-\frac{3}{4}\).

Therefore, the two true statements are:

  • You would use the inverse operation of square root.
  • There would be two solutions for (x; x = \frac{9}{12} = +\frac{3}{4} \text{ and } x = -\frac{9}{12} = -\frac{3}{4}.