Dave solved a quadratic equation. His work is shown below, with Step 111 missing. What could Dave have written as the result from Step 111? \begin{aligned} \dfrac{1}{3}(x+4)^2&=48 \\\\ &&\text{Step }1 \\\\ x+4&=\pm 12&\text{Step }2 \\\\ x=-16&\text{ or }x=8&\text{Step }3 \end{aligned} 3 1 (x+4) 2 x+4 x=−16 =48 =±12 or x=8 Step 1 Step 2 Step 3 Dave solved a quadratic equation. His work is shown below, with Step 111 missing. What could Dave have written as the result from Step 111? \begin{aligned} \dfrac{1}{3}(x+4)^2&=48 \\\\ &&\text{Step }1 \\\\ x+4&=\pm 12&\text{Step }2 \\\\ x=-16&\text{ or }x=8&\text{Step }3 \end{aligned} 3 1 (x+4) 2 x+4 x=−16 =48 =±12 or x=8 Step 1 Step 2 Step 3

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abidemiokin abidemiokin · Answer 1 · 2021-08-06T14:52:51+02:00

From the calculation above, Dave may have jumped the expression

[tex](x+4)^2= 144[/tex]

Given the expression solved by Dave from step 1 as;

[tex]\dfrac{1}{3}(x+4)^2&=48[/tex]

In order to determine what Dave would have written, we will solve the expression above as shown

[tex]\dfrac{1}{3}(x+4)^2&=48\\(x+4)^2 = 3*48\\(x+4)^2 = 144 ..............eqn \ 2\\[/tex]

Take the square root of both sides

[tex]\sqrt{(x+4)^2} =\pm\sqrt{144}\\x+4=\pm12[/tex]

Hence from the calculation above, Dave may have jumped the expression in equation 2 as shown below;

[tex](x+4)^2= 144[/tex]

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