How do you solve #sqrtx=8+x#?

2 Answers
Oct 16, 2015

Answer:

#x in O/#

Explanation:

You know that #x# must be positive because, for real numbers, you can only take the square root of positive numbers.

Start by squaring both sides of the equation to get rid of the square root

#(sqrt(x))^2 = (8 + x)^2#

#x = 64 + 16x + x^2#

Rearrange the ge all the terms on one side of the equation

#x^2 + 15x + 64 = 0#

Now, use the quadratic formula to find the two roots of this quadratic equation.

#x_(1,2) = (-15 +- sqrt(15^2 - 4 * 1 * 64))/(2 * 1)#

#x_(1,2) = (-15 +- sqrt(-31))/2#

Notice that the discriminant of the quadratic

#Delta = 15^2 - 4 * 1 * 64 = -31 <0#

is negative, which means that this quadratic equation has no real roots. It has, however, two complex roots that take the form

#x_(1,2) = (-15 +- isqrt(31))/2" "#, where #i^2 = -1#

This implies that the original equation has no real solutions either. In other words,

#sqrt(x) != 8 + x" ," AA x in RR" "#, or #" "x in O/#

Oct 16, 2015

Answer:

x= -15 #+-# #root#i#//#2

Explanation:

first square both sides,
you get,
x=#(8+x)^2#
x=64+#x^2# + 16x
that is, #x^2# + 15x +64 = 0
using -b #+-# #root##b^2#-4ac#//#2a
x= -15#+-# #root#-31#//#2
x= -15 #+-# i#root#31#//#2 (this is a complex number!)