Respuesta :
You can square both the sides and then use the formula for finding the roots of the obtained quadratic equation to find the solution needed.
The statements describing the solutions to the given equations are
- Option C: x = 3 is a true solution.
- Option E: The zeroes(also called solutions) of [tex]x^2 - 5x + 6 = 0[/tex] are possible solutions to the radical equation.
How to find the solutions to a quadratic equation?
Let the quadratic equation be [tex]ax^2 + bx + c = 0[/tex]
Then the solution to this quadratic equation is given as
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
What are extraneous solutions?
Those values of the variables who come out behaving like solutions but when tested, doesn't get proved as solutions to the given equation are called extraneous solutions.
Using that above conclusion to find the solutions to the given equation
Solutions to an equation means those values of the unknown for which the equation is true.
The given equation is
[tex]\sqrt{x-2} -4 = x - 6[/tex]
Converting that equation to quadratic form, we get
[tex]\sqrt{x-2} -4 = x - 6\\\\\sqrt{x-2} = x - 2\\[/tex]
From the obtained equation, we can see that x-2 is such a number whose square root is it itself. It is true for 2 numbers only.
First 1, and second 0.
Putting these values, we get
[tex]x-2 = 1\\x = 2+1 = 3\\\\x-2 = 0\\x = 2[/tex]
Thus, there are two solutions to the given equation. They are x = 2, and x = 3
Testing both the solutions by putting them in the given equation.
Case 1: x = 2
[tex]\sqrt{x-2} -4 = x - 6\\\sqrt{2-2} - 4 = 2 - 6\\0-4 = -4\\-4 = -4[/tex]
This is correct equality, thus, x = 2 is a correct solution.
Case 2: x = 3
[tex]\sqrt{x-2} -4 = x - 6\\\\\sqrt{3-2} - 4 = 3 - 6\\1 - 4 = -3\\-3 = -3[/tex]
This is correct equality, thus, x = 3 is a correct solution.
The zeroes(also called solutions) of [tex]x^2 - 5x + 6 = 0[/tex] is found as:
[tex]x^2 - 5x + 6 = 0\\x^2 - 3x - 2x +6 = 0\\(x-2)(x-3) = 0\\x = 2, x = 3[/tex]
Thus, the The zeroes(also called solutions) of [tex]x^2 - 5x + 6 = 0[/tex] are possible solutions to the radical equation given.
Thus,
- Option C: x = 3 is a true solution.
- Option E: The zeroes(also called solutions) of [tex]x^2 - 5x + 6 = 0[/tex] are possible solutions to the radical equation.
Learn more about extraneous solutions here:
https://brainly.com/question/26218294
Answer:
C and E are the correct answers.
Step-by-step explanation:
It's correct on Edgen. (2022)
