# Solve for x: square root of 7x+14 = x

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### 3 Answers

sqrt(7x+14) = x

First we will square both sides:

==> [sqrt(7x+14)]^2 = x^2

==> 7x + 14 = x^2

Now we will move all terms to one side:

==> 0 = x^2 - 7x - 14

==> x^2 -7x - 14 = 0

==> x1= [7 + sqrt(49-4*-14)]/2 = (7+ sqrt(105)/2

==> x2= (7- sqrt105)/2

Then the asnwer is:

**x= { (7+sqrt105)/2 , (7-sqrt105)/2 }**

Before solving the equation, we'll impose conditions of existence of the square root.

7x+14 >= 0

Wer'll subtract 14 both sides:

7x >= -14

We'll divide by 7: x >= -2

The interval of admissible solutions for the given equation is:

[-2 , +infinite)

Now, we'll solve the equation:

sqrt 7x+14 = x

We'll square raise both sides:

7x + 14 = x^2

We'll move all terms to one side and we'll use the symmetrical property:

x^2 - 7x - 14 = 0

We'll apply the quadratic formula:

x1 = [7+sqrt(49+56)]/2

x2 = [7-sqrt(49+56)]/2

since sqrt 105 = 10.24 approx

x1 = (7+10.24)/2

**x1 = 8.62 approx.**

x2 = (7-10.24)/2

**x2 = -1.62 approx.**

Since both values belong to the interval of admissible values, they are accepted as solutions of the given equation.

To solve square root of 7x+14 = x.

Solution:

(7x+14)^(1/2) - x = 0

Multiplying by (7x+14)^(1/2)+x both sides we get:

7x+14 - x^2 = 0

Multiplying by -1, we get:

x^2-7x-14 = 0. This is quadratic equation of the type ax^2+bx+c whose solutions are :

x1 = {-b+sqrt(b^2-4ac)}/2a or x2 = {-b-sqrt(b^2-4ac)}/2a.

Therefore a = 1 , b= 7 and c = -14 in x^2-7x-14 = 0.

So x1 = {7+sqrt(49+4*14)}/2 = (7+sqrt105)/2.

x2 = (7-sqrt105)/2 .