How do you solve #sqrtx=8+x#?
2 Answers
Explanation:
You know that
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/#
x= -15
Explanation:
first square both sides,
you get,
x=
x=64+
that is,
using -b
x= -15
x= -15