Respuesta :
We have that the equation is as follows:
[tex]\sqrt[]{x-2+8}=x[/tex]To solve this equation, we can proceed as follows:
1. Add the like terms in the radicand (-2 + 8 = 6):
[tex]\sqrt[]{x+6}=x[/tex]2. Square both sides of the equation:
[tex](\sqrt[]{x+6})^{^2}=x^2[/tex]3. We already know that the square root of a number is like raising that number to the fraction 1/2. Then, we have - using the power to a power rule:
[tex]((x+6)^{\frac{1}{2}})^2=(x+6)^{\frac{2}{2}}=(x+6)^1=x+6[/tex]4. Therefore, we have:
[tex]x+6=x^2[/tex]5. If we subtract x and 6 from both sides of the equation, we have:
[tex]x-x+6-6=x^2-x-6\Rightarrow0=x^2-x-6[/tex]6. Now, we have a second-degree (quadratic) polynomial that we can solve by factoring or by using the quadratic formula:
[tex]x^2-x-6=0[/tex]7. We can factor this polynomial if we find two numbers:
• a*b = -6
,• a + b = -1
Then, both numbers are:
• a = 2
,• b = -3
,• a*b = 2 * -3 = -6
,• a + b = 2 -3 = -1
Then, the above polynomial can be factored as follows:
[tex](x+2)(x-3)=0[/tex]8. The solutions for this equation are:
[tex]x+2=0\Rightarrow x=-2[/tex][tex]x-3=0\Rightarrow x=3[/tex]9. Now, to check these two solutions, we need to substitute these values into the original equation as follows:
For x = -2
[tex]\sqrt[]{-2-2+8}=-2\Rightarrow\sqrt[]{-4+8}=-2\Rightarrow\sqrt[]{4}=-2\Rightarrow2\ne-2[/tex]This is an extraneous solution. Although we got x = -2 as a solution to the resulting equation, it is not really a solution to the original equation. Therefore, x = -2 is an extraneous solution.
For x = 3
[tex]\sqrt[]{3-2+8}=3\Rightarrow\sqrt[]{1+8}=3\Rightarrow\sqrt[]{9}=3\Rightarrow3=3[/tex]Therefore, x = 3 is a solution to the original equation.
In summary, we have that the solution to the original equation:
[tex]\sqrt[]{x-2+8}=x[/tex]
is x = 3.
There is an extraneous solution, x = -2 (it is not an actual solution to the original equation).
