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 √)