Selesaikanlah dengan cara melengkapkan kuadrat sempurna! a. x²-2x-35=0 b. x²-10x+2=0 c. 2x²+6x+3=0

Jawaban:

Untuk metode melengkapkan kuadrat sempurna, harus ada:

1. Ruas kiri harus x² + bx

2. Gunakan rumus (½ · b)² kemudian ditambah ke kedua ruas

a. x² - 2x - 35 = 0

[tex]{x}^{2} - 2x - 35 = 0 \\ {x}^{2} - 2x = 35 \\ {x}^{2} - 2x + ( \frac{1}{2} b)^{2} = 35 + ( \frac{1}{2} b)^{2}\\ x^2 - 2x + ( \frac{1}{2} ( - 2) {)}^{2} = 35 + ( \frac{1}{2} ( - 2))^{2}\\ {x}^{2} - 2x + 1 = 35 + 1 \\ {(x - 1)}^{2} = 36 \\ \sqrt{ {(x - 1)}^{2} } = ±\sqrt{36} \\ \\ x - 1 = ±6 \\ \\ x = 6 + 1 ⟹x = 7 \\ x = - 6 + 1 ⟹x = - 5[/tex]

b. x² - 10x + 2 = 0

[tex]{x}^{2} - 10x + 2 = 0 \\ {x}^{2} - 10x = - 2 \\ {x}^{2} - 10x + ( \frac{1}{2} b {)}^{2} = - 2 + ( \frac{1}{2} b {)}^{2} \\ {x}^{2} - 10x + ( \frac{1}{2} ( - 10) {)}^{2} = - 2 + ( \frac{1}{2} ( - 10 {))}^{2} \\ {x}^{2} - 10x+ 25 = - 2 + 25 \\ (x - 5 {)}^{2} = 23 \\ \sqrt{ {(x - 5)}^{2} } = ± \sqrt{23} \\ x - 5 = ± \sqrt{23} \\ \\x - 5 = \sqrt{23} ⟹x = \sqrt{23} + 5 \\ x - 5 = - \sqrt{23} ⟹x = - \sqrt{23} + 5[/tex]

c. 2x² + 6x + 3 = 0

[tex]2 {x}^{2} + 6x + 3 = 0 \\ \frac{2 {x}^{2} }{2} + \frac{6x}{2} + \frac{3}{2} = 0 \\ {x}^{2} + 3x + \frac{3}{2} = 0 \\ {x}^{2} + 3x = - \frac{3}{2} \\ {x}^{2} + 3x + ( \frac{1}{2} b {))}^{2} = - \frac{3}{2} + ( \frac{1}{2} b) {)}^{2} \\ {x}^{2} + 3x + ( \frac{1}{2} (3) {)}^{2} = - \frac{3}{2} + ( \frac{1}{2} (3) {)}^{2} \\ {x}^{2} + 3x + \frac{9}{4} = - \frac{3}{2} + \frac{9}{4} \\ {x}^{2} + 3x + \frac{9}{4} = \frac{3}{4} \\ (x + \frac{3}{2} {)}^{2} = \frac{3}{4} \\ \sqrt{(x + \frac{3}{2} {)}^{2} } = ± \sqrt{ \frac{3}{4} } \\ x + \frac{3}{2} = ± \frac{ \sqrt{3} }{ \sqrt{2} } \\ x + \frac{3}{2} = ±\frac{ \sqrt{3} }{2} \\ \\ x + \frac{3}{2} = \frac{ \sqrt{3} }{2} ⟹x = \frac{ \sqrt{3} - 3 }{2} \\ x + \frac{3}{2} = - \frac{ \sqrt{3} }{2} ⟹x = - \frac{ \sqrt{3} - 3 }{2}

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