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

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Answer 1 · 2017-05-25T01:01:04

See a solution process below:

First, divide each side of the equation by #color(red)(7)# to isolate the #x^2# term while keeping the equation balanced:

#(7x^2)/color(red)(7) = -21/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x^2)/cancel(color(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: #{O/}#

Answer 2 · 2017-05-25T01:04:40

No real solutions but rather, two complex solution : #x=isqrt3,x=-isqrt3#

Divide #7# to both sides;

#cancel(7/7)x^2=-21/7#

#x^2=-3#

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

#x=+-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)*sqrt3, x=-sqrt(-1)*sqrt3#

*Recall that #sqrt-1=i#

Therefore we can simplify the expression as:

#x=isqrt3,x=-isqrt3# (This is our final answer)