Answer:
Third option: x=0 and x=16
Step-by-step explanation:
[tex]\sqrt{2x+4}-\sqrt{x}=2[/tex]
Isolating √(2x+4): Addind √x both sides of the equation:
[tex]\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}[/tex]
Squaring both sides of the equation:
[tex](\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}[/tex]
Simplifying on the left side, and applying on the right side the formula:
[tex](a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}[/tex]
[tex]2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x[/tex]
Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:
[tex]2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}[/tex]
Squaring both sides of the equation:
[tex](x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x[/tex]
This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:
[tex]x^{2}-16x=16x-16x\\ x^{2}-16x=0[/tex]
Factoring: Common factor x:
x (x-16)=0
Two solutions:
1) x=0
2) x-16=0
Solving for x: Adding 16 both sides of the equation:
x-16+16=0+16
x=16
Let's prove the solutions in the orignal equation:
1) x=0:
[tex]\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2[/tex]
x=0 is a solution
2) x=16
[tex]\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2[/tex]
x=16 is a solution
Then the solutions are x=0 and x=16
