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

##### 2 Answers

#### Answer:

#### Explanation:

You know that **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/#

#### Answer:

x= -15

#### Explanation:

first square both sides,

you get,

x=

x=64+

that is,

using -b

x= -15

x= -15