How to find the real solutions of the equation square root(2x+8)=x?

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tboneson1 eNotes educator| Certified Educator

Let's start by going back to some basics of solving single variable equations. One of the fundamental tools we use in algebra is the idea of inverse operations. Inverse means opposite, and operations means things like add, subtract, divide, multiply, take a root, or raise to a power. Inverse operations are operations that "undo" each other so to speak. Addition and subtraction undo each other; multiplying and dividing undo each other; Raising to a power and taking a root undo each other. Look at this simple example: x + 2 = 5

To solve this problem we need to "undo" everything that is happening to x. If you examine the left side of the equation, you will see that x is being added by 2. The inverse/opposite of this would be subtracting 2, and that is what we will do to "undo" what has happened to x. We are allowed to do this, or any other operation we choose, so long as we do the same thing equally on both sides of the equal sign.

Think of an equation like a balance scale. In a balance scale you have two bowls attached to a lever of some kind, and the two bowls are balanced on opposite ends of the lever. If you put 1lb in the left bowl, it will drop down lower than the right bowl. To "rebalance" the scale, we need to put the same amount of weight (1lb) in the right bowl, and the two bowls will again be balanced. Equations are like this. If we subtract 2 from one side of the equal sign, then we must also subtract 2 from the other side of the equal sign to maintain the balance of the equation.

x + 2 = 5

-2 -2

x = 3

So now on to your problem: square root(2x+8)=x

As we did in the simple example of x + 2 = 5, we want to undo any operations that have been done to x, so the first step is to remove the radical (square root box) that has the x trapped inside. To do this, we need the operation that is the opposite of taking a square root, which is raising to the second power or squaring. So we are going to raise everything on both sides of the equal sign to the second power. On the left side of the equal sign, what will happen is that the square root and the squaring(raising to the second power) will cancel each other out.

This is the same as the idea from the simpler example of x + 2 = 5. When we subtracted to from the left side of that problem, the +2 and -2 cancelled each other out because they were inverse/opposite operations.

Now back to our problem. We've raised the left side to the second power, so remember our scale, we need to do the same thing on the right side. After squaring both sides we should have:

2x+8=x squared

Now we have a quadratic equation which we can solve either by factoring or using the quadratic formula. We'll factor this one.

First we have to get everything on one side and set the problem equal to zero, which we can do by subtracting 2x and subtracting 8 from both sides. That will leave us 0=x squared -2x-8. We will factor the quadratic term, x squared, which is x times x and those will become the first factor in each of our binomial factors:

0=(x )(x )

Now we must find the factors of our constant -8 that have a sum/difference of -2. These are -4 and +2, because (-4)(2) is -8, but -4+2 is -2. We now put those in our binomial factors:


Lastly, we use the zero product property to find our solutions. Take the two factors and set them each equal to zero and solve:

x-4=0 and x+2=0

+4 +4 -2 -2

so x=4 or x=-2, these are your two real solutions.

giorgiana1976 | Student

First, we'll impose constraints of existence of the square root:

`2x + 8 gt= 0`

`` `x + 4 gt= 0`

`` `x gt= -4`

Now, we'll raise to square both side to remove the square root:

`2x + 8 = x^2`

We'll subtract 2x + 8 both sides and we'll use symmetrical property:

`x^2 - 2x - 8 = 0`

We'll apply quadratic formula to determine the roots x1 and x2:

x1 = `(2+sqrt(4 + 32))/2`

x1 = `(2+sqrt36)/2`

x1 = `(2+6)/2`

x1  =4

x2  = `(2-6)/2`

x2 = -2

Since both values of x are included within interval `[-4;+oo), ` then the solutions of the equation are x1 = -2 and x2 = 4.