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Match the imaginary solutions with the correct simplified forms:

Match The Imaginary Solutions With The Correct Simplified Forms class=

Sagot :

Answer:

See the picture below.

Step-by-step explanation:

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Answer:

[tex]\sqrt{-63}=3i\sqrt{7}[/tex]

[tex]\sqrt{-156}=2i\sqrt{39}[/tex]

[tex]6\sqrt{-12}=12i\sqrt{3}[/tex]

[tex]-5\sqrt{-75}=-25i\sqrt{3}[/tex]

Step-by-step explanation:

An imaginary number is a non-zero real number multiplied by the imaginary unit [tex]i[/tex].

[tex]\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Imaginary Number Rule}\end{center}\\\begin{center}$\sqrt{-1}=i$\end{center}\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\begin{center}\underline{Radical Rule}\end{center}\\\begin{center}$\sqrt{a \cdot b}=\sqrt{a} \sqrt{b}$\end{center}\end{minipage}}[/tex]

[tex]\begin{aligned}\sqrt{-63} & = \sqrt{9 \cdot -1 \cdot 7}\\& = \sqrt{9}\sqrt{-1}\sqrt{7}\\& = 3i\sqrt{7}\end{aligned}[/tex]

[tex]\begin{aligned}\sqrt{-156} & = \sqrt{4 \cdot -1 \cdot -39}\\& = \sqrt{4} \sqrt{-1} \sqrt{39}\\& = 2i\sqrt{39}\end{aligned}[/tex]

[tex]\begin{aligned}6 \sqrt{-12} & = 6 \sqrt{4 \cdot -1 \cdot -3} \\& = 6 \sqrt{4} \sqrt{-1} \sqrt{3} \\& = 6 \cdot 2 i \sqrt{3} \\& = 12i \sqrt{3}\end{aligned}[/tex]

[tex]\begin{aligned}-5 \sqrt{-75} & = -5 \sqrt{25 \cdot -1 \cdot -3} \\& = -5 \sqrt{25} \sqrt{ -1 } \sqrt{3} \\& = -5 \cdot 5i \sqrt{3} \\& = -25i \sqrt{3}\end{aligned}[/tex]

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