x, squared, plus, 4, equals, 0
x
2
+4=
0
Solve the problem just like the example below
⬇️
Solve for all real values of x.
x, squared, minus, 16, equals, 0
x
2
−16=
0
plus, 16, plus, 16
+16=
+16
x, squared, equals, 16
x
2
=
16
square root of, x, squared, end square root, equals, plus or minus, square root of, 16, end square root
x
2
=
±
16
x, equals, plus or minus, 4
x=
±4
Solutions: , x, equals, 4, comma, x, equals, minus, 4
Solutions: x=4,x=−4
1 answer
Solve for all real values of x.
\( x^2 + 4 = 0 \)
Subtract 4 from both sides:
\( x^2 + 4 - 4 = 0 - 4 \)
\( x^2 = -4 \)
Take the square root of both sides:
\( \sqrt{x^2} = \pm \sqrt{-4} \)
\( x = \pm \sqrt{-4} \)
Since \( \sqrt{-4} = 2i \) (where \( i \) is the imaginary unit), we rewrite it as:
\( x = \pm 2i \)
Solutions: \( x = 2i, x = -2i \)
Since there are no real solutions, we conclude:
There are no real values of \( x \) that satisfy the equation.
