Respuesta :

Answer:

B

Step-by-step explanation:

using Pythagoras' identity in the right triangle

the square on the hypotenuse is equal to the sum of the squares on the other 2 sides , then

(x + 1)² + (x + 3)² = (x + 5)² ← expand all factors using FOIL

x² + 2x + 1 + x² + 6x + 9 = x² + 10x + 25 , that is

2x² + 8x + 10 = x² + 10x + 25 ← subtract x² + 10x + 25 from both sides

x² - 2x - 15 = 0 ← in standard form

(x - 5)(x + 3) = 0 ← in factored form

equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

x + 3 = 0 ⇒ x = - 3

however, x > 0 , then x = 5

HayhayHay010101

Answer:

B. 5

Step-by-step explanation:

[tex] \sf{C = \sqrt{ A {}^{2} + B {}^{2} }}[/tex]

[tex]\sf{x + 5 = \sqrt{ {(x + 3)}^{2} + {(x + 1)}^{2} }}[/tex]

[tex]\sf{x + 5 = \sqrt{ {(x {}^{2} + 6x + 9)} + {(x {}^{2} + 2x + 1)}}}[/tex]

[tex]\sf{x + 5 = \sqrt{ {(2x {}^{2} + 8x + 10)}}}[/tex]

[tex]\sf{(x + 5) {}^{2} = { {2x {}^{2} + 8x + 10}}}[/tex]

[tex]\sf{ {x}^{2} + 10x + 25= { {2x {}^{2} + 8x + 10}}}[/tex]

[tex]\sf{0= { {2x {}^{2} \red{ { - x}^{2} } + 8x \red{ - 10x} + 10 \red{ - 25}}}}[/tex]

[tex]\sf{0= x^{2} } - 2x { - 15}[/tex]

[tex]\sf{0= x^{2} } - 2x { - 15}[/tex]

[tex]\sf{0= (x - 5)(x + 3)}[/tex]

[tex] \sf{x_{1} - 5 = 0 }[/tex]

[tex] \sf{x_{1} = 0 + 5 }[/tex]

[tex] \sf{x_{1} = \boxed{5 }}[/tex]

[tex] \sf{x_{2} + 3 = 0 }[/tex]

[tex] \sf{x_{2} = 0 - 3 }[/tex]

[tex] \sf{x_{2} = \boxed{- 3 }}[/tex]

Option B. 5

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