Question

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

Answer 1

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

Answer 2

Answer:

$x=\frac{-5+\sqrt{65}}{2},\frac{-5-\sqrt{65}}{2}$, or x=1.531 and x=-6.531.

Step-by-step 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:

$\sqrt{x+14}+2=x\\\\\sqrt{x+14}+2-2=x-2\\\\\sqrt{x+14}=x-2$

To cancel the radical, we square both sides:

$(\sqrt{x+14})^2=(x-2)^2\\\\x+14=(x-2)(x-2)\\\\x+14=x(x)-2(x)-2(x)-2(-2)\\\\x+14=x^2-2x-2x--4\\\\x+14=x^2-4x--4\\\\x+14=x^2-4x+4$

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):

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\=\frac{-5\pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}\\\\=\frac{-5\pm \sqrt{25--40}}{2}\\\\=\frac{-5\pm \sqrt{65}}{2}\\\\=\frac{-5+\sqrt{65}}{2},\frac{-5-\sqrt{65}}{2}$

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