QUESTION IMAGE
Question
solve: ( x^2 + 3 = 19 )
a. ( pm 4 )
b. ( pm sqrt{4} )
c. ( pm 16 )
d. ( pm 22 )
e. ( pm sqrt{22} )
Step1: Isolate the \(x^2\) term
Subtract 3 from both sides of the equation \(x^2 + 3 = 19\).
We get \(x^2=19 - 3\).
Simplifying the right - hand side, \(19-3 = 16\), so \(x^2 = 16\).
Step2: Solve for \(x\)
Take the square root of both sides of the equation \(x^2=16\).
Recall that if \(y^2 = a\) (\(a\geq0\)), then \(y=\pm\sqrt{a}\).
Here, \(a = 16\), so \(x=\pm\sqrt{16}\).
Since \(\sqrt{16}=4\), we have \(x = \pm4\).
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A. $\pm4$