Verified Answer :

naǫ

a.
\(x^2=49 \\ \sqrt{x^2}=\sqrt{49} \\ |x|=7 \\ x=-7 \ \lor \ x=7 \\ \boxed{x=-7 \hbox{ or } x=7}\)

b.
\(x^2-7=29 \ \ \ |+7 \\ x^2=36 \\ \sqrt{x^2}=\sqrt{36} \\ |x|=6 \\ x=-6 \ \lor \ x=6 \\ \boxed{x=-6 \hbox{ or } x=6}\)

c.
\(4x^2=64 \ \ \ |\div 4 \\ x^2=16 \\ \sqrt{x^2}=\sqrt{16} \\ |x|=4 \\ x=-4 \ \lor \ x=4 \\ \boxed{x=-4 \hbox{ or } x=4}\)

lavortmeyer

(a)
X²=49
x=7, -7
(b)
x²-7=29
x²=36
x=6, -6
(c)
4x²=64
x²=16
x=4, -4

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