Question

tentukan akar akar penyelesaian dari bentuk 2x² + 5x + 3 = 0 dengan melengkapi kuadrat sempurna​​

Answer 1

Jawaban:

Penyelesaian :

2x² + 5x + 3 = 0

• Kuadrat sempurna

[tex]\begin{aligned}\sf \underline{{2x}^{2} +5x + 3} &= \sf 0 \: \: \to( \div 2) \\ \sf {x}^{2} + \frac{5}{2} x + \frac{3}{2} &= \sf 0 \\ \sf {x}^{2} + \frac{5}{2}x&= \sf - \frac{3}{2} \\ \sf {x}^{2} + \frac{5}{2} x + (\frac{5}{4}) {}^{2}&= \sf - \frac{3}{2} + ( \frac{5}{4} ) {}^{2} \\ \sf (x + \frac{5}{4}) {}^{2}&= \sf - \frac{3}{2} + \frac{25}{16} \\ \sf (x + \frac{5}{4} ) {}^{2} &= \sf - \frac{24 + 25}{16} \\ \sf (x + \frac{5}{4} ) {}^{2}&= \sf \frac{1}{16} \\ \sf (x + \frac{5}{4} )&= \sf \sqrt{ \frac{1}{16} } \\ \sf (x + \frac{5}{4}) &= \sf \frac{1}{4} \\ \\ \to \sf (x + \frac{5}{4} )&= \sf \frac{1}{4} \\ \sf x&= \sf \frac{1}{4} - \frac{5}{4} \\ \sf x &= \sf - \frac{4}{4} \\ \sf x&= \sf \red{ - 1} \\ \\ \sf \to (x + \frac{5}{4}) &= \sf - \frac{1}{4} \\ \sf x&= \sf - \frac{1}{4} - \frac{5}{4} \\ \sf x &= \sf - \frac{6}{4} \\ \sf x &= \sf \red{ - \frac{3}{2} }\end{aligned}[/tex]

[tex] \sf HP = \{ \red{-1 ,-\frac{3}{2}} \}[/tex]

'조슈아' (Svt)

Answer 2

1. 2x^2 + 5x + 3 = 0 (bagi ÷2)

x^2 + 5/2x + 3/2 = 0

5/2(1/2) = 5/4

(x + 5/4)^2 = x^2 + 10x/2 + 25/16

(x + 5/4)^2 = x^2 + 5x + 25/16

x^2 + 5/2x = -3/2

(x + 5/4)^2 = -3/2 + 25/16

(x + 5/4)^2 = 1/16

x + 5/4 = √1/√16

x + 5/4 = ± 1/4

x + 5/4 = 1/4

x = 1/4 - 5/4

x = -4/4

x = -1

x + 5/4 = -1/4

x = -1/4 - 5/4

x = -6/4

x = -3/2

HP = {-1, -3/2}