Penjelasan dengan langkah-langkah:
bc = √(ab² + ac²)
(2x + 1) = √(x² + (2x)²)
(2x + 1)² = x² + 4x²
4x² + 4x + 1 = 5x²
4x² - 5x² + 4x + 1 = 0
-x² + 4x + 1 = 0
x² - 4x - 1 = 0
x1,x2 = -b ± √(b² - 4ac)/2a
x1,x2 = -(-4) ± √((-4)² - 4 . 1 . (-1))/2(1)
x1,x2 = 4 ± √(16 + 4)/2
x1,x2 = (4 ± √20)/2
x1,x2 = (4 ± √(4 x 5))/2
x1,x2 = (4 ± 2√5/2)
x1,x2 = 2(2 ± √5)/2
x1,x2 = 2 ± √5
utk 2 - √5
= 2 - 2,236
= -0,236 (tdk memenuhi)
utk 2 + √5
= 2 + 2,236
= 4,236 cm
sehingga nilai x = (2 + √5) cm
pembuktian
(2x + 1)² = x² + 2x²
(2(2 + √5) + 1)² = (2 + √5)² + (2(2 + √5))²
(4 + 2√5 + 1)² = (4 + 4√5 + 5) + (4 + 2√5)²
(5 + 2√5)² = (9 + 4√5) + (16 + 16√5 + 20)
(25 + 20√5 + 20) = (9 + 4√5) + (36 + 16√5)
45 + 20√5 = 45 + 20√5 (terbukti √)
[tex]x = - 2 \: ± \: \sqrt{3} [/tex]
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Rumus Teorema Pythagoras : a² = b² + c²
Keterangan :
a² = b² + c²
(2x + 1)² = (2x)² + (x)²
4x² + 2x + 2x + 1 = 4x² + x²
4x² - 4x² + 2x + 2x + 1 = 0
4x + 1 + x² = 0
x² + 4x + 1 = 0
[tex]x = \frac{ - b \: ± \: \sqrt{ {b}^{2} } - 4ac }{2a} [/tex]
[tex]x \: = \: \frac{ - 4 \: ± \: \sqrt{ {4}^{2} \: - 4 \: \times \: 1\: \times \: 1 } }{2 \: \times \: 1} [/tex]
[tex]x = \frac{ - 4 \: ± \: \sqrt{16 - 4 \times 1 \times 1} }{2 \times 1}[/tex]
[tex]x = \frac{ - 4 \: ± \: \sqrt{12} }{2} [/tex]
[tex]x= \frac{ - 4 \: ± \: 2 \sqrt{3} }{2} [/tex]
[tex]x = \frac{ - 4 \: + \: 2 \sqrt{3} }{2} [/tex]
[tex]x = \frac{ - 4 \: - 2 \sqrt{3} }{2} [/tex]
[tex]x = - 2 \: + \: \sqrt{3} [/tex]
[tex]x = - 2 \: - \: \sqrt{3} [/tex]
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Penjelasan dengan langkah-langkah:
bc = √(ab² + ac²)
(2x + 1) = √(x² + (2x)²)
(2x + 1)² = x² + 4x²
4x² + 4x + 1 = 5x²
4x² - 5x² + 4x + 1 = 0
-x² + 4x + 1 = 0
x² - 4x - 1 = 0
x1,x2 = -b ± √(b² - 4ac)/2a
x1,x2 = -(-4) ± √((-4)² - 4 . 1 . (-1))/2(1)
x1,x2 = 4 ± √(16 + 4)/2
x1,x2 = (4 ± √20)/2
x1,x2 = (4 ± √(4 x 5))/2
x1,x2 = (4 ± 2√5/2)
x1,x2 = 2(2 ± √5)/2
x1,x2 = 2 ± √5
utk 2 - √5
= 2 - 2,236
= -0,236 (tdk memenuhi)
utk 2 + √5
= 2 + 2,236
= 4,236 cm
sehingga nilai x = (2 + √5) cm
pembuktian
(2x + 1)² = x² + 2x²
(2(2 + √5) + 1)² = (2 + √5)² + (2(2 + √5))²
(4 + 2√5 + 1)² = (4 + 4√5 + 5) + (4 + 2√5)²
(5 + 2√5)² = (9 + 4√5) + (16 + 16√5 + 20)
(25 + 20√5 + 20) = (9 + 4√5) + (36 + 16√5)
45 + 20√5 = 45 + 20√5 (terbukti √)
Verified answer
[tex]x = - 2 \: ± \: \sqrt{3} [/tex]
____________________________
Rumus Teorema Pythagoras : a² = b² + c²
Keterangan :
Penyelesaian :
a² = b² + c²
(2x + 1)² = (2x)² + (x)²
4x² + 2x + 2x + 1 = 4x² + x²
4x² - 4x² + 2x + 2x + 1 = 0
4x + 1 + x² = 0
x² + 4x + 1 = 0
Rumus Kuadrat :
[tex]x = \frac{ - b \: ± \: \sqrt{ {b}^{2} } - 4ac }{2a} [/tex]
x² + 4x + 1 = 0
[tex]x \: = \: \frac{ - 4 \: ± \: \sqrt{ {4}^{2} \: - 4 \: \times \: 1\: \times \: 1 } }{2 \: \times \: 1} [/tex]
[tex]x = \frac{ - 4 \: ± \: \sqrt{16 - 4 \times 1 \times 1} }{2 \times 1}[/tex]
[tex]x = \frac{ - 4 \: ± \: \sqrt{12} }{2} [/tex]
[tex]x= \frac{ - 4 \: ± \: 2 \sqrt{3} }{2} [/tex]
[tex]x = \frac{ - 4 \: + \: 2 \sqrt{3} }{2} [/tex]
[tex]x = \frac{ - 4 \: - 2 \sqrt{3} }{2} [/tex]
[tex]x = - 2 \: + \: \sqrt{3} [/tex]
[tex]x = - 2 \: - \: \sqrt{3} [/tex]
[tex]x = - 2 \: ± \: \sqrt{3} [/tex]
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