[tex]\bf a)\,x^2=144 \implies x=\pm \sqrt{144} = \pm \,12[/tex]
[tex]\bf \implies x \in \{ 12 ;- 12\}[/tex]
[tex]\bf b) \, x^2=256 \implies x= \pm \sqrt{256} = \pm \, 16[/tex]
[tex]\bf \implies x \in \{16;-16\}[/tex]
[tex]\bf c)\, x^2=400 \implies x= \pm \sqrt{400}=\pm \, 20[/tex]
[tex]\bf \implies x \in \{20;-20\}[/tex]
[tex]\bf d)\, x^2=576 \implies x= \pm \sqrt{576}= \pm \, 24[/tex]
[tex]\bf \implies x \in \{24;-24\}[/tex]
[tex]\bf e)\, x^2=625 \implies x= \pm \sqrt{625} = \pm \, 25[/tex]
[tex]\bf \implies x \in \{25;-25\}[/tex]
[tex]\bf f)\, 3x^2=300 \, |:3 \implies x^2=100[/tex]
[tex]\bf \implies x= \pm \sqrt {100} = \pm \, 10 \implies x \in \{10;-10\}[/tex]
[tex]\bf g)\,4x^2=324\, |:4 \implies x^2=81[/tex]
[tex]\bf \implies x = \pm \sqrt{81}=\pm\, 9 \implies x \in \{9;-9\}[/tex]
[tex]\bf h)\, 7x^2=6300\, |:7 \implies x^2=900[/tex]
[tex]\bf \implies x = \pm\sqrt{900}= \pm \, 30 \implies x \in \{30;-30\}[/tex]
[tex]\bf i) \, 10x^2=490 \, |:10\implies x^2=49[/tex]
[tex]\bf \implies x= \pm \sqrt{49} = \pm \, 7 \implies x \in \{7;-7\}[/tex]
[tex]\triangle \triangle \triangle[/tex]
[tex]\bf a)\, x^2=0 \implies x=0[/tex]
[tex]\bf b)\, x^2=-2 (nr\: negativ) \implies x \notin R[/tex]
[tex]\bf c)\, x^2=16 \implies x=\pm \sqrt{16}=\pm\, 4[/tex]
[tex]\bf \implies x\in \{4;-4\}[/tex]
[tex]\bf d)\, 4x^2=64 \, |:4 \implies x^2=16[/tex]
[tex]\bf \implies x=\pm \sqrt{16} = \pm \, 4 \implies x \in \{4;-4\}[/tex]
[tex]\bf e)\, 3x^2=108 \, |:3 \implies x^2=36[/tex]
[tex]\bf \implies x=\pm \sqrt{36}=\pm\, 6 \implies x \in \{6;-6\}[/tex]
[tex]\bf f)\,x^2+4=5 \implies x^2=5-4=1[/tex]
[tex]\bf \implies x= \pm \sqrt{1} \implies x \in \{1;-1\}[/tex]
[tex]\bf g)\, 3x^2+2=1^2+(-1)^2 \implies 3x^2+2=1+1=2[/tex]
[tex]\bf \implies 3x^2+2=2_{ /-2} \implies 3x^2=0\, |:3 \implies x=0[/tex]
[tex]\bf h)\, x^2+1=(\sqrt{3})^2\implies x^2+1=3_{/-1}[/tex]
[tex]\bf \implies x^2=2 \implies x=\pm \sqrt{2} \implies x\in \{\sqrt{2};-\sqrt{2}\}[/tex]
[tex]\bf i)\, x^2=\dfrac{4}{25} \implies x = \pm \sqrt{\dfrac{4}{25} }[/tex]
[tex]\bf \implies x= \pm \dfrac{2}{5} \implies x\in \bigg\{\dfrac{2}{5} ;-\dfrac{2}{5}\bigg\}[/tex]
Bafta! :)
