Question

65pts

Answer 1

Answer:

Part 1)

$\sqrt{x+2}=x$  ......[1]

Equation 1 has an extraneous solution.

$\sqrt{3x+1}=4$  ......[2]

Equation 2 does not have extraneous solution.

Part 2)

• $\sqrt{x+2}=x$ has an extraneous solution which is x = -1.
• the equation $\sqrt{3x+1}=4$ does not have an extraneous solution.

Part 3)

• The first equation $\sqrt{x+2}=x$ has an extraneous solution. The reason is that x = -1 does not satisfy the equation.
• The second equation $\sqrt{3x+1}=4$ 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

$\sqrt{x+2}=x$  ......[1]

Equation 1 has an extraneous solution.

$\sqrt{3x+1}=4$  ......[2]

Equation 2 does not have extraneous solution.

Part 2)

Solving Equation [1]

$\sqrt{x+2}=x$  ......[1]

$\mathrm{Square\:both\:sides}$

$\left(\sqrt{x+2}\right)^2=x^2$

$x+2=x^2$

$x^2-x-2=0$

$(x-2)(x+1)=0$

$x=2,\:x=-1$

$\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}\sqrt{x+2}=x$

Plugging x = 2

$\sqrt{2+2}=2$

$2 = 2$

2 is true.

Plugging x = -1

$\sqrt{\left(-1\right)+2}=-1$

$1 = -1$

-1 does not satisfy the solution. Therefore, -1 is an extraneous root.

Therefore, $\sqrt{x+2}=x$ has an extraneous solution which is x = -1.

Solving Equation [2]

$\sqrt{3x+1}=4$  ......[2]

$Square\:both\:sides$

$\left(\sqrt{3x+1}\right)^2=4^2$

$3x+1=16$

$3x=15$

$x=5$

$\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}\sqrt{3x+1}=4$

Plugging x = 5

$\sqrt{3\cdot \:5+1}=4$

$4=4$

As x = 5 satisfies the equation. So, the equation $\sqrt{3x+1}=4$ does not have an extraneous solution.

Hence, x = 5 is the solution of $\sqrt{3x+1}=4$.

Part 3)

The first equation $\sqrt{x+2}=x$ has an extraneous solution. The reason is that x = -1 does not satisfy the equation.

Check this verification:

Plugging x = -1

$\sqrt{\left(-1\right)+2}=-1$

$1 = -1$

Hence, $\sqrt{x+2}=x$ has an extraneous solution which is x = -1.

The second equation $\sqrt{3x+1}=4$ does not have an extraneous. The reason is that x = 5 satisfies the equation.

Check this verification:

Plugging x = 5

$\sqrt{3\cdot \:5+1}=4$

$4=4$

Hence, $\sqrt{3x+1}=4$ does not have an extraneous solution.

Keywords: radical equation, extraneous solution

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