How do you solve #x^2+8=0# using the quadratic formula?

Question

Eva Adamson · Accepted Answer

#x=2sqrt(2)color(white)("x")icolor(white)("XXX")orcolor(white)("XXX")x=-2sqrt(2)color(white)("x")i#

(see below for use of the quadratic formula)

Standard quadratic form for an equation is: #color(white)("XXX")color(red)ax^2+color(blue)bx+color(magenta)c=0# and in this form the solutions are given by #color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(magenta)c))/(2color(red)a)# Expressing the given equation into standard form: #x^2+8=0# #color(white)("XXX")rarr color(red)1x^2+color(blue)0 * x +color(magenta)8 = 0# with solutions: #color(white)("XXX")x=(-color(blue)0+-sqrt(color(blue)0^2-4 * color(red)1 * color(magenta)8))/(2 * color(red)1)# #color(white)("XXX")=+-sqrt(-32)/2=+-(4sqrt(2)color(white)("x")i)/2=+-2sqrt(2)color(white)("x")i#

Julia Adams · Answer

undefined To solve the equation $x^2 + 8 = 0$ using the quadratic formula, follow these steps:

1. Identify the coefficients in the equation:
   $a = 1$,
   $b = 0$,
   $c = 8$.

2. Substitute the values of $a$, $b$, and $c$ into the quadratic formula:
   $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.

3. Plug in the values:
   $x = \frac{{-0 \pm \sqrt{{0^2 - 4 \cdot 1 \cdot 8}}}}{{2 \cdot 1}}$.

4. Simplify inside the square root:
   $x = \frac{{\pm \sqrt{{-32}}}}{2}$.

5. Rewrite the square root of a negative number as $i$, the imaginary unit:
   $x = \frac{{\pm \sqrt{{32}} \cdot i}}{2}$.

6. Simplify the square root of 32:
   $x = \frac{{\pm 4\sqrt{2} \cdot i}}{2}$.

7. Reduce the expression:
   $x = \pm 2\sqrt{2} \cdot i$.

So, the solutions to the equation $x^2 + 8 = 0$ are $x = 2\sqrt{2} \cdot i$ and $x = -2\sqrt{2} \cdot i$.