# How do you solve 7x^2=-21?

May 25, 2017

See a solution process below:

#### Explanation:

First, divide each side of the equation by $\textcolor{red}{7}$ to isolate the ${x}^{2}$ term while keeping the equation balanced:

$\frac{7 {x}^{2}}{\textcolor{red}{7}} = - \frac{21}{\textcolor{red}{7}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} {x}^{2}}{\cancel{\textcolor{red}{7}}} = - 3$

${x}^{2} = - 3$

Next, we would take the square root of each side of the equation to solve for $x$ while keeping the equation balanced.

However, there is no Real solution for the square root of a negative number, in this case the $\sqrt{- 3}$

Therefore, there is no real solution or the solution set is the empty or null set: $\left\{\emptyset\right\}$

May 25, 2017

No real solutions but rather, two complex solution : $x = i \sqrt{3} , x = - i \sqrt{3}$

#### Explanation:

Divide $7$ to both sides;

$\cancel{\frac{7}{7}} {x}^{2} = - \frac{21}{7}$

${x}^{2} = - 3$

sqrt(x^2)=sqrt(-3 (Apply the square root property)

$x = \pm \sqrt{- 3}$

There are no "real solutions" since the number inside the radical is negative but there are two "complex" solutions:

We can rewrite the expression above as:

$x = \sqrt{- 1} \cdot \sqrt{3} , x = - \sqrt{- 1} \cdot \sqrt{3}$

*Recall that $\sqrt{-} 1 = i$

Therefore we can simplify the expression as:

$x = i \sqrt{3} , x = - i \sqrt{3}$ (This is our final answer)