Question↷
Which statement best reflects the solution(s) of the equation?
[tex] \small \sqrt{2x - 1} - x + 2 = 0[/tex]
Answer↷
There is only one solution: x = 5 .
The solution x = 1 is an extraneous solution.
Solution↷
[tex] \small \sqrt{2x - 1} - x + 2 = 0[/tex]
- Taking (-x+2) to the other side
[tex] \small \sqrt{2x - 1} = x - 2 [/tex]
- squaring both of the sides
[tex] \small 2x - 1 = {(x - 2 )}^{2} [/tex]
- expanding the sqared binomial as ㅤㅤㅤㅤ(a-b)²= a²- 2ab + b²
[tex] \small 2x - 1 = {x}^{2} - 2 \times x \times 2 + {2}^{2} [/tex]
[tex] \small 2x - 1 = {x}^{2} - 4x + 4[/tex]
[tex] \small {x}^{2} - 4x - 2x + 4 + 1 = 0\: [/tex]
[tex] \small {x}^{2} - 6x + 5 = 0\: [/tex]
- using splitting middle term method
[tex] \small {x}^{2} - 5x - x+ 5 = 0\: [/tex]
[tex] \small x(x- 5) - (x - 5 )= 0\: [/tex]
[tex] \small (x- 1)(x - 5 )[/tex]
so the root would either be 5 or 1
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putting the value of x as 1,
[tex] \small \sqrt{2 \times 1 - 1} - 1 + 2 = 0[/tex]
[tex] \small 1 - 1 + 2 = 0[/tex]
[tex] \small 3 - 1≠ 0[/tex]
hence , it's not the true solution of the equation
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putting the value of x as 5,
[tex] \small \sqrt{2 \times 5 - 1} - 5 + 2 = 0[/tex]
[tex] \small 3 - 5 + 2 = 0[/tex]
[tex] \small 5 - 5 = 0[/tex]
[tex] \small 0 = 0[/tex]
hence ,it's the true solution of the equation
