How do you solve the quadratic #7n^2+6=-40# using any method?

1 Answer
May 3, 2018

Answer:

#x = (+- isqrt322)/7#

Explanation:

#7n^2 + 6 = -40#

To use the quadratic formula to find the zeroes, we need to make sure the equation is written in the form #color(red)(a)x^2 + color(magenta)(b)x + color(blue)(c) = 0#.

To do so, we need to add #color(orange)40# to both sides of the equation:
#7n^2 + 6 quadcolor(orange)(+quad40) = -40 quadcolor(orange)(+quad40)#

#7n^2 + 46 = 0#

So we know that:
#color(red)(a = 7)#

#color(magenta)(b = 0)#

#color(blue)(c = 46)#

The quadratic formula is #x = (-color(magenta)(b) +- sqrt(color(magenta)(b)^2 - 4color(red)(a)color(blue)(c)))/(2color(red)(a))#.

Now we can plug in the values for #color(red)(a)#, #color(magenta)(b)#, and #color(blue)(c)# into the quadratic formula:

#x = (-color(magenta)(0) +- sqrt((color(magenta)(0))^2 - 4(color(red)(7))(color(blue)(46))))/(2(color(red)(7)))#

Simplify:
#x = (+- sqrt(-1288))/14#

The solution is imaginary since we cannot take the square root of a negative number.

However, we know that #i#, or imaginary, is equal to #sqrt(-1)#. Therefore, we can factor out a #-1# from the square root and make it an #i#:

#x = (+-isqrt(1288))/14#

#x = (+- 2isqrt(322))/14#

#x = (+- isqrt322)/7#

This is the same thing as:
#x = (isqrt322)/7# and #x = (-isqrt322)/7#
because #+-# means "plus or minus."

To clarify, this doesn't mean that there is a zero or root. This is an imaginary solution, meaning that there are no zeros. To prove this, let's look at the graph of this equation:
enter image source here
(desmos.com)

As you can see, there are no zeroes. The vertex starts above the #x#-axis.

Hope this helps!