Task 2 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. Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

a(x+b)+c=d that has an extraneous solution

a(x+b)+c=d that does not have an extraneous solution

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

part a)

for extraneous solution

1 ⋅ sqrt(x+2) + 3 = 0

for non extraneous solution

1⋅ sqrt(x+2)  + 3 = 6

part b) solve each equation

1⋅sqrt(x+2)+3=0

x+sqrt(x+2)=−3

square both sides

(sqrt(x+2))^ 2 = (−3)^2

x+2=9

x=9−2

x=7

do you see why its extraneous

Step-by-step explanation:

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.