Rezolvati ecuatile a 3x²=12 c - x²+5=0b 2x²-6=0d x²+4=0.... Idea question from @cri231

July 2022 1 44 Report

Rezolvati ecuatile

a 3x²=12
c - x²+5=0
b 2x²-6=0
d x²+4=0

$3 {x}^{2} = 12 \\ {x}^{2} = 12 \div 3 \\ {x}^{2} = 4 \\ x = \sqrt{4} \\ x = + - 2$

x={ -2;2}

$- {x}^{2} + 5 = 0 \\ - {x}^{2} = - 5 \\ {x}^{2} = 5 \\ x = \sqrt{5}$

$2 {x}^{2} - 6 = 0 \\ 2 {x}^{2} = 6 \\ {x}^{2} = 6 \div 2 \\ {x}^{2} = 3 \\ x = \sqrt{3}$

${x}^{2} + 4 = 0 \\ {x}^{2} = - 4fals$

un număr la puterea a doua nu poate să fie negativ

…ᘛ⁐̤ᕐᐷ…ᘛ⁐̤ᕐᐷฅ^•ﻌ•^ฅ

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