What is the solution set for #2x^2 + 4x +10 = 0#?

2 Answers
Aug 26, 2015

Answer:

There are no Real solutions for the given equation.

Explanation:

We can see that there are no Real solutions by checking the discriminant
#color(white)("XXX")b^2-4ac#

#color(white)("XXX")= 16 - 80 < 0 color(white)("XX")rarrcolor(white)("XX")no Real roots

or
If we look at the graph for the expression, we can see that it does not cross the X-axis and therefore is not equal to zero at any values for #x#:
graph{2x^2+4x+10 [-10, 10, -5, 5]}

Aug 26, 2015

Answer:

#x_(1,2) = (-1 +- 4i)/2#

Explanation:

For a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

you can determine its roots by using the quadratic formula

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a))#

Now, you can divide all the terms by #2# to make the calculations easier

#(color(red)(cancel(color(black)(2)))x^2)/color(red)(cancel(color(black)(2))) + (4/2)x + 10/2 = 0#

#x^2 + 2x + 5 = 0#

For this quadratic, you have #a=1#, #b=2#, and #c=5#, which means that the two roots will be

#x_(1,2) = (-1 +- sqrt(2^2 - 4 * 1 * 5))/(2 * 1)#

Notice that the determinant, #Delta#, which is the name given to the expression that's under the square root, , is negative.

#Delta = b^2 - 4ac#

#Delta = 2^2 - 4 * 1 * 5 = -16#

For real numbers, you cannot take the square root of a negative number, which means that the quadratic equation has no real solutions.

Its graph will not intercept the #x#-axis. However, it will have two distinct complex roots.

#x_(1,2) = (-1 +- sqrt(-16))/2#

#x_(1,2) = (-1 +- (i^2 * 16))/2 = (-1 +- i * sqrt(16))/2#

#x_(1,2) = (-1 +- 4i)/2#

The two roots will thus be

#x_1 = (-1 + 4i)/2" "# and #" "x_2 = (-1 - 4i)/2#