Question

Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model.
a√x+b+c=d

Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous.

Part 3. Explain why the first equation has an extraneous solution and the second does not.

Answer 1

Answer:

1)a) $\sqrt{2x+3}=x$ b)$\sqrt{x+5}=3$ 2) Solving and Checking on section 1 3) Commonly when we square both sides and there's an x on the right side outside the radical equation, an extraneous solution arises.

Step-by-step explanation:

1 &2) An extraneous solution does not satisfy an equation. So let's evaluate:

$a=1,b=3,c=0, d=x}$

$\sqrt{2x+3}=x$

$\sqrt{2x+3}=x\\\left ( \sqrt{2x+3} \right )^{2}=\left (x\right)^{2}\\2x+3=x^{2}\\-x^{2}+2x+3=0\:*(-1)\\x^{2}-2x-3=0\\(x-3)(x+1)=0\\x'=-1, x''=3 \\S=\left \{ -1,3 \right \}$

Checking for extraneous solution:

$\sqrt{2(-1)+3}=-1\\\sqrt{3-2}=-1\\\sqrt{1}\neq-1\\1\neq -1\\\therefore -1 \:is \:an\:extraneous\:solution$

$\sqrt{2(3)+3}=3\\\sqrt{6+3}=3\\\sqrt{9}=3\\3= 3\\\therefore 3 \:satisfies\:the\:equation.$

1 and 2) Another radical equation:

$a=1, b=5,c=0, d=3$

$\sqrt{x+5}=3\\\left ( \sqrt{x+5}\right )^{2}=3^{2}\\x+5=9\\x=9-5\\x=4\\S=\left \{ 4\right \}$

Checking:

$\sqrt{x+5}=3\\\left ( \sqrt{4+5} \right )^{2}=3^{2}\\4+5=9\\9=9$

4 satisfies the equation.

3) Because when we change the radical equation into quadratic one, and both sides are squared, very often the second solution does not satisfy the equation. That happened when we squared x.

Answer 2

A+x=bxc
bc=aa 
bc=a
esa es la respuesta final