Question

Identify all of the following solutions of square root of x plus 14 end root plus 2 equals x.

Answer 1

`\bf √(x+14)+2=x\implies √(x+14)=x-2\impliedby \textit{squaring both sides} \\\\\\ (√(x+14))^2=(x-2)^2\implies x+14=x^2-4x+4 \\\\\\ 0=x^2-5x-10\impliedby \textit{now, using the quadratic formula} \\\\\\ x=\cfrac{5\pm√((-5)^2-4(1)(-10))}{2(1)}\implies x=\cfrac{5\pm√(25+40)}{2} \\\\\\ x=\cfrac{5\pm √(65)}{2}\implies x= \begin{cases} \cfrac{5+ √(65)}{2}\\\\ \cfrac{5- √(65)}{2} \end{cases}`

Answer 2

`x=(-5+√(65))/(2),(-5-√(65))/(2)`, or x=1.531 and x=-6.531.

Explanation:

The first thing we will do is isolate the radical expression on the left hand side. To do this, we subtract 2 from each side:

`√(x+14)+2=x\\\\√(x+14)+2-2=x-2\\\\√(x+14)=x-2`

To cancel the radical, we square both sides:

`(√(x+14))^2=(x-2)^2\\\\x+14=(x-2)(x-2)\\\\x+14=x(x)-2(x)-2(x)-2(-2)\\\\x+14=x^2-2x-2x--4\\\\x+14=x^2-4x--4\\\\x+14=x^2-4x+4`

Now we will cancel the x on the left hand side by subtraction:

x+14-x = x²-4x+4-x

14 = x²-5x+4

Make the equation equal 0 by subtracting 14 from each side:

14-14 = x²-5x+4-14

0 = x²-5x-10

Now we use the quadratic formula (in our equation, a=1, b=-5 and c=-10):

`x=(-b\pm √(b^2-4ac))/(2a)\\\\=(-5\pm √((-5)^2-4(1)(-10)))/(2(1))\\\\=(-5\pm √(25--40))/(2)\\\\=(-5\pm √(65))/(2)\\\\=(-5+√(65))/(2),(-5-√(65))/(2)`

When you evaluate the square root, you get the answers 1.531 and -6.531.