Question

tentukan akar akar penyelesaian dari bentuk x² + 2x - 8 = 0 dengan melengkapi kuadrat sempurna​

Answer 1

Jawaban:

Penyelesaian :

x² + 2x - 8 = 0

• Kuadrat sempurna

[tex]\begin{aligned}\sf {x}^{2} +2x -8 &= \sf 0 \\ \sf {x}^{2} +2x&= \sf 8 \\ \sf {x}^{2} 2x + ( \frac{2}{2}) {}^{2} &= \sf 8 + ( \frac{2}{2} ) {}^{2} \\ \sf (x + 1) {}^{2} &= \sf 8+ 1² \\ \sf (x +1) {}^{2}&= \sf 8 + 1 \\ \sf ( x +1 ) {}^{2} &= \sf 9\\ \sf x +&= \sf \sqrt{ 9} \\ \sf x +1&= \sf ± 3 \\ \\ \sf \to (x +1) &= \sf 3\\ \sf x&= \sf 3-1 \\ \sf x&= \sf \red{ 2 } \\ \\ \sf \to (x + 1) &= \sf - 3\\ \sf x&= \sf - 3-1 \\ \sf x&= \sf \red{-4}\end{aligned}[/tex]

[tex] \sf HP = \{ \red{-4,2} \}[/tex]

'조슈아' (Svt)

Answer 2

Jawaban:

x+4=0

x-2=0

Penjelasan dengan langkah-langkah:

x² + 2x - 8 = 0

x²+4x-2x-8=0

x×(x+4)-2(x+4)=0

(x+4)×(x-2)=0

x+4=0

x-2=0