Using thedefinition of a square root, explain why is not a real number.
There is no real number such that

Ask by Flynn Tran. in the United States

Mar 28,2025

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Using thedefinition of a square root, explain why $$ \sqrt{-25} $$ is not a real number. There is no real number $$ b $$ such that $$ b $$

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By definition, the square root of a number $$ a $$ is a value $$ b $$ such that $$ b^2 = a $$. This means that for any real number $$ a $$, if $$ a $$ is non-negative, there exists a real number $$ b $$ such that $$ b^2 = a $$.
In the case of $$ \sqrt{-25} $$, we are looking for a real number $$ b $$ such that $$ b^2 = -25 $$. However, the square of any real number is always non-negative (i.e., $$ b^2 \geq 0 $$). Therefore, there cannot be a real number $$ b $$ that satisfies this equation since $$ -25 $$ is negative.
Thus, $$ \sqrt{-25} $$ is not a real number because there is no real number $$ b $$ such that $$ b^2 = -25 $$.

Using thedefinition of a square root, explain why is not a real number.
There is no real number such that

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Using thedefinition of a square root, explain why is not a real number. There is no real number such that