Question

2026sp-math-147-001-16w: precalculus
homework: section 3.1 complex numbers
question 3, 3.1.9
express in terms of $i$.
$sqrt{-252}$
$sqrt{-252} = square$
(simplify your answer. type your answer in the form a
question list
question 1
question 2
question 3
question 4
help me solve this

Explanation:

Step1: Split negative radicand

$\sqrt{-252} = \sqrt{-1 \times 252}$

Step2: Separate square roots

$\sqrt{-1 \times 252} = \sqrt{-1} \times \sqrt{252}$

Step3: Simplify $\sqrt{-1}$ and factor 252

$\sqrt{-1}=i$; $\sqrt{252}=\sqrt{36 \times 7}$

Step4: Simplify $\sqrt{36 \times 7}$

$\sqrt{36 \times 7}=\sqrt{36} \times \sqrt{7}=6\sqrt{7}$

Step5: Combine terms

$i \times 6\sqrt{7} = 6i\sqrt{7}$

Answer:

$6i\sqrt{7}$