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prisoha [69]
3 years ago
11

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

Mathematics
2 answers:
AfilCa [17]3 years ago
7 0

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.

BabaBlast [244]3 years ago
5 0
\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}
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