To solve the equation [tex]\( y + 2 \sqrt{y-8} = 0 \)[/tex], let's go through it step by step.
1. Isolate the square root term:
[tex]\( y + 2 \sqrt{y-8} = 0 \)[/tex]
Subtract [tex]\( y \)[/tex] from both sides:
[tex]\[
2 \sqrt{y-8} = -y
\][/tex]
2. Square both sides to get rid of the square root (but remember that squaring can introduce extraneous solutions, so you must check potential solutions in the original equation):
[tex]\[
(2 \sqrt{y-8})^2 = (-y)^2
\][/tex]
This simplifies to:
[tex]\[
4(y-8) = y^2
\][/tex]
3. Expand and rearrange the quadratic equation:
[tex]\[
4y - 32 = y^2
\][/tex]
Moving all terms to one side gives:
[tex]\[
y^2 - 4y + 32 = 0
\][/tex]
4. Solve the quadratic equation:
For the quadratic equation [tex]\( y^2 - 4y + 32 = 0 \)[/tex], we can use the quadratic formula, [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 32 \)[/tex].
[tex]\[
y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1}
\][/tex]
Simplifying inside the square root:
[tex]\[
y = \frac{4 \pm \sqrt{16 - 128}}{2}
\][/tex]
[tex]\[
y = \frac{4 \pm \sqrt{-112}}{2}
\][/tex]
Since the discriminant ([tex]\( \sqrt{-112} \)[/tex]) is negative, the quadratic equation has no real roots.
5. Conclusion:
Since there are no real roots, the equation [tex]\( y + 2 \sqrt{y-8} = 0 \)[/tex] has no real solutions.
Thus, the solution set for the equation [tex]\( y + 2 \sqrt{y-8} = 0 \)[/tex] is empty:
[tex]\[
[]
\][/tex]
