Asked by Desiree LeBoeuf on May 21, 2024

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Solve the equation by using the Square Root Property. 9u2+30=09 u ^ { 2 } + 30 = 09u2+30=0

A) u=±30u = \pm \sqrt { 30 }u=±30
B) u=±309iu = \pm \frac { \sqrt { 30 } } { 9 } iu=±930i
C) u=±303u = \pm \frac { \sqrt { 30 } } { 3 }u=±330
D) u=±309u = \pm \frac { \sqrt { 30 } } { 9 }u=±930
E) u=±303iu = \pm \frac { \sqrt { 30 } } { 3 } iu=±330i

Square Root Property

The square root property states that if \(x^2 = a\), then \(x = \pm\sqrt{a}\), assuming \(a\) is non-negative.

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TH
Tameka HarleyMay 26, 2024
Final Answer :
E
Explanation :
To use the Square Root Property, we need to isolate the variable, which means we need to move the constant term to the other side of the equation:
9u2=−30u2=−309u2=−103\begin{align*} 9u^2 &= -30 \\u^2 &= -\frac{30}{9} \\u^2 &= -\frac{10}{3}\end{align*}9u2u2u2=30=930=310
However, since we cannot take the square root of a negative number in the real number system, we have to use imaginary numbers. The square root of $-1$ is denoted by $i$, so we have:
u=±−103=±−103=±−103⋅33=±−309=±309⋅−1=±303i\begin{align*}u &= \pm \sqrt{-\frac{10}{3}} \\&= \pm \sqrt{\frac{-10}{3}} \\&= \pm \sqrt{\frac{-10}{3} \cdot \frac{3}{3}} \\&= \pm \sqrt{\frac{-30}{9}} \\&= \pm \frac{\sqrt{30}}{\sqrt{9}} \cdot \sqrt{-1} \\&= \pm \frac{\sqrt{30}}{3} i \\\end{align*}u=±310=±310=±31033=±930=±9301=±330i
Therefore, the answer is E.