Answer:
Part 1)
......[1]
Equation 1 has an extraneous solution.
......[2]
Equation 2 does not have extraneous solution.
Part 2)
has an extraneous solution which is x = -1.- the equation
does not have an extraneous solution.
Part 3)
- The first equation
has an extraneous solution. The reason is that x = -1 does not satisfy the equation. - The second equation
does not have an extraneous solution. The reason is that x = 5 satisfies the equation.
Step-by-step explanation:
Part 1)
Creating Two Radical Equations
Let us consider two radical equations
......[1]
Equation 1 has an extraneous solution.
......[2]
Equation 2 does not have extraneous solution.
Part 2)
<u><em>Solving Equation [1]</em></u>
......[1]
![\mathrm{Square\:both\:sides}](https://tex.z-dn.net/?f=%5Cmathrm%7BSquare%5C%3Aboth%5C%3Asides%7D)
![\left(\sqrt{x+2}\right)^2=x^2](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%7Bx%2B2%7D%5Cright%29%5E2%3Dx%5E2)
![x+2=x^2](https://tex.z-dn.net/?f=x%2B2%3Dx%5E2)
![x^2-x-2=0](https://tex.z-dn.net/?f=x%5E2-x-2%3D0)
![(x-2)(x+1)=0](https://tex.z-dn.net/?f=%28x-2%29%28x%2B1%29%3D0)
![x=2,\:x=-1](https://tex.z-dn.net/?f=x%3D2%2C%5C%3Ax%3D-1)
![\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}\sqrt{x+2}=x](https://tex.z-dn.net/?f=%5Cmathrm%7BCheck%5C%3Athe%5C%3Asolutions%5C%3Aby%5C%3Aplugging%5C%3Athem%5C%3Ainto%5C%3A%7D%5Csqrt%7Bx%2B2%7D%3Dx)
Plugging x = 2
![\sqrt{2+2}=2](https://tex.z-dn.net/?f=%5Csqrt%7B2%2B2%7D%3D2)
![2 = 2](https://tex.z-dn.net/?f=2%20%3D%202)
2 is true.
Plugging x = -1
![\sqrt{\left(-1\right)+2}=-1](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%28-1%5Cright%29%2B2%7D%3D-1)
![1 = -1](https://tex.z-dn.net/?f=1%20%3D%20-1)
-1 does not satisfy the solution. Therefore, -1 is an extraneous root.
Therefore,
has an extraneous solution which is x = -1.
<u><em>Solving Equation [2]</em></u>
......[2]
![Square\:both\:sides](https://tex.z-dn.net/?f=Square%5C%3Aboth%5C%3Asides)
![\left(\sqrt{3x+1}\right)^2=4^2](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%7B3x%2B1%7D%5Cright%29%5E2%3D4%5E2)
![3x+1=16](https://tex.z-dn.net/?f=3x%2B1%3D16)
![3x=15](https://tex.z-dn.net/?f=3x%3D15)
![x=5](https://tex.z-dn.net/?f=x%3D5)
![\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}\sqrt{3x+1}=4](https://tex.z-dn.net/?f=%5Cmathrm%7BCheck%5C%3Athe%5C%3Asolutions%5C%3Aby%5C%3Aplugging%5C%3Athem%5C%3Ainto%5C%3A%7D%5Csqrt%7B3x%2B1%7D%3D4)
Plugging x = 5
![\sqrt{3\cdot \:5+1}=4](https://tex.z-dn.net/?f=%5Csqrt%7B3%5Ccdot%20%5C%3A5%2B1%7D%3D4)
![4=4](https://tex.z-dn.net/?f=4%3D4)
As x = 5 satisfies the equation. So, the equation
does not have an extraneous solution.
Hence, x = 5 is the solution of
.
Part 3)
The first equation
has an extraneous solution. The reason is that x = -1 does not satisfy the equation.
Check this verification:
Plugging x = -1
![\sqrt{\left(-1\right)+2}=-1](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cleft%28-1%5Cright%29%2B2%7D%3D-1)
![1 = -1](https://tex.z-dn.net/?f=1%20%3D%20-1)
Hence,
has an extraneous solution which is x = -1.
The second equation
does not have an extraneous. The reason is that x = 5 satisfies the equation.
Check this verification:
Plugging x = 5
![\sqrt{3\cdot \:5+1}=4](https://tex.z-dn.net/?f=%5Csqrt%7B3%5Ccdot%20%5C%3A5%2B1%7D%3D4)
![4=4](https://tex.z-dn.net/?f=4%3D4)
Hence,
does not have an extraneous solution.
<em>Keywords: radical equation, extraneous solution</em>
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