Answer:
A and B
Step-by-step explanation:
we would like to solve the following equation:
[tex] \rm\displaystyle 5 - 2x = \sqrt{ {2x}^{2} + x - 1 } - x[/tex]
to do so isolate -x to the left hand side and change its sign:
[tex] \rm\displaystyle 5 - 2x + x = \sqrt{ {2x}^{2} + x - 1 } [/tex]
simplify addition:
[tex] \rm\displaystyle 5 - x = \sqrt{ {2x}^{2} + x - 1 } [/tex]
square both sides:
[tex] \rm\displaystyle (5 - x {)}^{2} = (\sqrt{ {2x}^{2} + x - 1 } {)}^{2} [/tex]
simplify square of the right hand side:
[tex] \rm\displaystyle (5 - x {)}^{2} = {2x}^{2} + x - 1 [/tex]
use (a-b)²=a²-2ab+b² to expand the left hand side:
[tex] \rm\displaystyle {x}^{2} - 10x + 25= {2x}^{2} + x - 1 [/tex]
swap the equation:
[tex] \rm\displaystyle {2x}^{2} + x - 1 = {x}^{2} - 10x + 25[/tex]
isolate the right hand side expression to the left hand side and change every sign:
[tex] \rm\displaystyle {2x}^{2} + x - 1 - {x}^{2} + 10x - 25 = 0[/tex]
simplify:
[tex] \rm\displaystyle {x}^{2} + 11x - 26= 0[/tex]
rewrite the middle term as 13x-2x:
[tex] \rm\displaystyle {x}^{2} + 13x - 2x - 26= 0[/tex]
factor out x:
[tex] \rm\displaystyle x({x}^{} + 13)- 2x - 26= 0[/tex]
factor out -2:
[tex] \rm\displaystyle x({x}^{} + 13)- 2(x + 13)= 0[/tex]
group:
[tex] \rm\displaystyle (x- 2)(x + 13)= 0[/tex]
by Zero product property we obtain:
[tex] \displaystyle \begin{cases}x- 2 = 0\\ x + 13= 0 \end{cases}[/tex]
solve for x:
[tex] \displaystyle \begin{cases}x = 2\\ x = - 13 \end{cases}[/tex]
to check for extraneous solutions we can define the domain of equation recall that a square root of a function is always greater than or equal to 0 therefore
[tex] \rm\displaystyle 5 - x \geq0[/tex]
solve the inequality for x:
[tex] \rm\displaystyle x \leqslant 5[/tex]
since 2 and -13 is less than 5 both solutions are valid for x hence,
[tex] \displaystyle \begin{cases}x _{1} = 2\\ x_{2} = - 13 \end{cases}[/tex]
and we're done!
