Suma y resta de cantidades imaginarias puras.

. √-4 + √-16

Se reducen a la forma de una cantidad real (a) multiplicada por √-1 y luego se reducen como radicales semejantes.

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Ejemplos:



a) Simplificar √-4 + √-9

> Convirtiendo a la forma a√-1

√-4 = √[(4)(-1)] = (√2²)(√-1) = 2√-1
√-9 = √[(9)(-1)] = (√3²)(√-1) = 3√-1
> Reduciendo como radicales semejantes:
2√-1 + 3√-1

= (2+3)√-1

= 5√-1

= 5i Solución.

b) Simplificar 2√-36 - √-25 + √-12

> Convirtiendo a la forma a√-1:

2√-36 = 2[(√36)(-1)] = 2(√6²)(√-1) = 2(6)√-1 = 12√-1

√-25 = [(√25)(-1)] = (√5²)(√-1) = 5(√-1) = 5√-1
√-12 = [(√12)(-1)] = [√2²(3)](√-1) = 2(√3)(√-1)

> Reduciendo los radicales semejantes:

12√-1 – 5√-1 + 2(√3)(√-1)

= (12-5+2√3)(√-1)

= (7+2√3)(√-1)
= (7+2√3)i Solución.

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Ejercicio 254

Simplificar:

1) √-4 + √-16

√-4 = √[(4)(-1)] = (√2²)(√-1) = 2(√-1)
√-16 = √[(16)(-1)] = (√4²)(√-1) = 4(√-1)
→ 2(√-1) + 4(√-1)
= (2+4)√-1
= 6√-1
= 6i Solución.

2) √-25 + √-81 - √-49
√-25 = √[(25)(-1)] = (√5²)(√-1) = 5(√-1)
√-81 = √[(81)(-1)] = (√9²)(√-1) = 9(√-1)
√-49 = √[(49)(-1)] = (√7²)(√-1) = 7(√-1)
→ 5(√-1) + 9(√-1) – 7(√-1)
= (5+9-7)√-1
= 7√-1
= 7i Solución.

3) 2√-9 + 3√-100
2√-9 = 2√[(9)(-1)] = 2√(3²)(√-1) = 2(3)√-1 = 6√-1
3√-100 = 3√[(100)(-1)] = 3√(10²)(√-1) = 3(10)√-1 = 30√-1
→ 6√-1 + 30√-1
= (6+30)√-1
= 36√-1
= 36i Solución.

6) √-18 + √-8 + 2√-50
√-18 = √[(18)(-1)] = (√3²)(2)(√-1) = 3(√2)(√-1)
√-8 = √[(8)(-1)] = (√2²)(2)(√-1) = 2(√2)(√-1)
2√-50 = 2√[(50)(-1)] = 2(√5²)(2)(√-1) = 2(5)(√2)(√-1)
→ 3(√2)(√-1) + 2(√2)(√-1) + 10(√2)(√-1)
= (3+2+10)√2√-1
= 15√2√-1
= 10√2i Solución.

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