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Use the square root property to determine all real solutions for each of the following equations. \( \begin{array}{l}3 a^{2}-132=0 \\ \\ 2 z^{2}+16=0 \\ z= \\ \text { Give exact solutions (don't use decimals), and separate multiple solutions with commas. If there are no rea } \\ \text { solutions, type DNE. }\end{array} \)

Ask by Spencer Gibson. in the United States
Mar 09,2025

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Answer

The solutions are: - For \(3a^2 - 132 = 0\): \(a = 2\sqrt{11},\, -2\sqrt{11}\) - For \(2z^2 + 16 = 0\): DNE

Solution

1. For the equation \(3a^2 - 132 = 0\): - Add \(132\) to both sides: \[ 3a^2 = 132 \] - Divide both sides by \(3\): \[ a^2 = \frac{132}{3} = 44 \] - Take the square root of both sides: \[ a = \sqrt{44} \quad \text{or} \quad a = -\sqrt{44} \] - Simplify \(\sqrt{44}\) since \(44 = 4 \times 11\): \[ \sqrt{44} = \sqrt{4 \times 11} = 2\sqrt{11} \] - The exact solutions are: \[ a = 2\sqrt{11},\, -2\sqrt{11} \] 2. For the equation \(2z^2 + 16 = 0\): - Subtract \(16\) from both sides: \[ 2z^2 = -16 \] - Divide both sides by \(2\): \[ z^2 = -8 \] - Since the square of a real number cannot be negative, there are no real solutions. - The result is: \[ \text{DNE} \]

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To solve the equations using the square root property, let's go through each one step-by-step. 1. For the equation \( 3a^{2} - 132 = 0 \): - First, add 132 to both sides: \[ 3a^{2} = 132 \] - Next, divide both sides by 3: \[ a^{2} = 44 \] - Now, take the square root of both sides: \[ a = \pm \sqrt{44} \] - Simplifying \( \sqrt{44} \) gives: \[ a = \pm 2\sqrt{11} \] - Thus, the solutions are: \[ a = 2\sqrt{11}, -2\sqrt{11} \] 2. For the equation \( 2z^{2} + 16 = 0 \): - First, subtract 16 from both sides: \[ 2z^{2} = -16 \] - Next, divide both sides by 2: \[ z^{2} = -8 \] - Since the square of a real number cannot be negative, there are no real solutions. Therefore, we write: \[ z = \text{DNE} \] Combining results: The solutions are \( \begin{array}{l} a = 2\sqrt{11}, -2\sqrt{11} \\ z = \text{DNE} \end{array} \)

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