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3 years ago

HURRYYYY

Mathematics

2 answers:

GarryVolchara [31]3 years ago

6 0

Answer:

There are no real number solutions.

Step-by-step explanation:

$2x^{2} +8=0\\2x^{2} =-8\\x^{2} =-4\\x=\sqrt{-4}$

Real numbers are a whole range of numbers including: Natural Numbers(1;2;3...), Whole Numbers(0;1;2...), Integers(...;-2;-1;0;1;2;...), Rational Numbers(-0.6; -0.2; 0 ;1.5 etc.) and Irrational Numbers( $\sqrt{2}$ ; $\sqrt{3}$ ;π; etc.). All these number comprise Real Numbers.

Imaginary Numbers are the square roots of negative Integers, which is the case for this question.

Radda [10]3 years ago

5 0

There is no help me to do this sorry

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