Respuesta :
The given equation is
[tex]\sqrt[3]{1+x+\sqrt{1+2x}}=2[/tex]First, we need to elevate each side to the third power.
[tex]\begin{gathered} (\sqrt[3]{1+x+\sqrt{1+2x}})^3=(2)^3 \\ 1+x+\sqrt{1+2x}=8 \end{gathered}[/tex]Second, subtract x and 1 on both sides.
[tex]\begin{gathered} 1+x+\sqrt{1+2x}-x-1=8-x-1 \\ \sqrt{1+2x}=7-x \end{gathered}[/tex]Third, we elevate the equation to the square power to eliminate the root
[tex]\begin{gathered} (\sqrt{1+2x})^2=(7-x)^2 \\ 1+2x=(7-x)^2 \end{gathered}[/tex]Now, we use the formula to solve the squared binomial.
[tex](a-b)=a^2-2ab+b^2[/tex][tex]\begin{gathered} 1+2x=7^2-2(7)(x)+x^2 \\ 1+2x=49-14x+x^2 \end{gathered}[/tex]Now, we solve this quadratic equation
[tex]\begin{gathered} 0=49-14x+x^2-2x-1 \\ x^2-16x-48=0 \end{gathered}[/tex]We need to find two number which product is 48 and which difference is 16. Those numbers are 12 and 4, we write them down as factors.
[tex]x^2-16x-48=(x-12)(x+4)[/tex]So, the possible solutions are
[tex]\begin{gathered} x-12=0\rightarrow x=12 \\ x+4=0\rightarrow x=-4 \end{gathered}[/tex]However, we need to verify each solution to ensure that each of them satisfies the given equation. We just need to evaluate it with those two solutions.
[tex]\begin{gathered} \sqrt[3]{1+x+\sqrt{1+2x}}=2\rightarrow\sqrt[3]{1+12+\sqrt{1+2(12)}}=2 \\ \sqrt[3]{13+\sqrt{1+24}}=2 \\ \sqrt[3]{13+\sqrt{25}}=2 \\ \sqrt[3]{13+5}=2 \\ \sqrt[3]{18}=2 \\ 2.62=2 \end{gathered}[/tex]As you can observe, the solution 12 doesn't satisfy the given equation.
