How do you solve #5x^2+64=0#?

1 Answer
Oct 20, 2016

Answer:

This can be solved using the quadratic formula.

Explanation:

While there are many ways to solve quadratic equations like this one, the quadratic formula is one of the simplest.

The quadratic formula looks like this:
#x=(-b+-sqrt(b^2-4ac))/2a#

To use this equation, we simply plug in the values given in the problem. To find these values, we must acknowledge that a quadratic equation looks like this:
#y=ax^2+bx+c#

Once we know this, we can determine that our equation has an a value of 5, because it is multiplied by #x^2#. The b value must be zero, because none is given. If a number (x) is multiplied by zero, it will disappear. Finally, the c value must be 64, because it is not multiplied by any x value.
Now, we can plug in the values of a, b, and c.
#x=(-0+-sqrt(0^2-4(5)(64)))/(2(5))#

We simplify to get:
#x=(+-sqrt(-1280))/10#

The #+-# gives us two values for x. We solve for #+# and #-# separately. Because the value under the radical is negative, the answer will involve #i#, which is the variable given to the #sqrt(-1)#.
When we simplify, the two x values are:
#x=-(8sqrt(5)i)/5#
#x=(8sqrt(5)i)/5#