How do you simplify $(2 + \sqrt{- 25}) + (- 8 - \sqrt{- 9})$ $(2+ \sqrt { - 25} ) + ( - 8- \sqrt { - 9} )$ ?

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How do you simplify $(2 + \sqrt{- 25}) + (- 8 - \sqrt{- 9})$ $(2+ \sqrt { - 25} ) + ( - 8- \sqrt { - 9} )$ ?

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Answer 1 · 2018-07-25T05:36:37

$\Rightarrow - 6 + 2 i$ $color(purple)(=> -6 + 2i$

Explanation:

$(2 + \sqrt{-} 25) + (- 8 - \sqrt{- 9}$ $(2 + sqrt-25 ) + (-8 - sqrt(-9)$

$\Rightarrow (2 + \sqrt{25 i^{2}}) + (- 8 - \sqrt{9 i^{2}}), as i^{2} = - 1$ $=> (2 + sqrt (25 i^2) ) + (-8 - sqrt(9 i^2)), " as " i^2 = -1$

$\Rightarrow (2 + 5 i) + (- 8 - 3 i)$ $=> (2 + 5i) + (-8 - 3i)$

$\Rightarrow (2 - 8) + (5 i - 3 i)$ $=> (2 - 8) + (5i - 3i)$

$\Rightarrow - 6 + 2 i$ $color(purple)(=> -6 + 2i$

Answer 2 · 2018-07-25T08:02:45

$- 6 + 2 i$ $-6+2i$

Explanation:

$note that \sqrt{- 1} = i$ $"note that "sqrt(-1)=i$

$∙ x \sqrt{a} \times \sqrt{b} \Leftrightarrow \sqrt{a b}$ $•color(white)(x)sqrtaxxsqrtbhArrsqrt(ab)$

$\sqrt{- 25} = \sqrt{25 \times - 1} = \sqrt{25} \times \sqrt{- 1} = 5 i$ $sqrt(-25)=sqrt(25xx-1)=sqrt25xxsqrt(-1)=5i$

$\sqrt{- 9} = \sqrt{9 \times - 1} = \sqrt{9} \times \sqrt{- 1} = 3 i$ $sqrt(-9)=sqrt(9xx-1)=sqrt9xxsqrt(-1)=3i$

$2 + \sqrt{- 25} - 8 - \sqrt{- 9}$ $2+sqrt(-25)-8-sqrt(-9)$

$= 2 + 5 i - 8 - 3 i$ $=2+5i-8-3i$

$= (2 - 8) + (5 i - 3 i)$ $=(2-8)+(5i-3i)$

$= - 6 + 2 i$ $=-6+2i$

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How do you simplify $(2 + \sqrt{- 25}) + (- 8 - \sqrt{- 9})$ $(2+ \sqrt { - 25} ) + ( - 8- \sqrt { - 9} )$ ?

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How do you simplify (2+√−25)+(−8−√−9)(2+ \sqrt { - 25} ) + ( - 8- \sqrt { - 9} )?

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How do you simplify $(2 + \sqrt{- 25}) + (- 8 - \sqrt{- 9})$ $(2+ \sqrt { - 25} ) + ( - 8- \sqrt { - 9} )$ ?