Question

HELP PLEASEEE!!!! flvs

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

@PurplePunch

Answer 2

I have to leave now, if you can help me please private message me, thanks !! @PurplePunch @jollyjolly0

Answer 3

https://www.desmos.com/calculator/71vm6md7kl

Try messing with the constants on that. (i had to correct the equation they gave you, it should be d*x, not d). You can see that i moved the d*x term to the other side, so the solutions are when a (sqrt(x+b)) + c - dx = 0
I plotted it against y so you can see a graph, which means the solutions are when y = 0 (on the x axis)

So yea, just mess with the constants on that until you find an equation that intersects the x axis (y = 0) twice, that does not have an extraneous solution. Then find one that only intersects the x axis (y = 0) once, that has one extraneous solution.

show your work in solving them (square both sides and use quadratic equation), and at the end, plug both of the solutions back into the original equation. If one of the answers does not make both sides of the equation equal, it is an extraneous solution, and you can discard it.

Good luck!

Answer 4

@jollyjolly0 Thank but I understood absolutely nothing ):

Answer 5

if you take the sample they gave you,
a√(x+b)+c=d*x
like i said, I replaced d with d*x because I believe you mistyped it.

rearranging, you get
a√(x+b)+c -d*x = 0

Remember, that a, b, c and d are all just numbers. The difference between a,b,c,d and x is that when we actually try to solve this, we will put in actual numbers (like 1 or 3.2 or 9/2) for a,b,c,d but we won't for x. After we have replaced a,b,c,d with numbers, we will then try to find the value for x that makes
a√(x+b)+c -d*x = 0

In the calculator I linked, it plots
y = a√(x+b)+c -d*x
and there are sliders for a,b,c and d so you can pick what number thy are.

but as you saw above, the actual solutions for x are when
a√(x+b)+c -d*x = 0
by comparing that to
a√(x+b)+c -d*x = y

we can see that the solutions on the graph are when y = 0

so if you open the graph and move around the sliders for a,b,c and d, you can see that sometimes the curve intersects the x axis in one place, and sometimes it does in 2 places. When it intersects in one place, that means you have 1 solution.
to find a radical expression that has one solution, mess with the sliders until the curve only intersects the x axis at 1 point. Then, write out
a√(x+b)+c -d*x = 0
take your values for a, b, c and d that were in the calculator sliders and put them into your equation. So say in the calculator , a = 5, b = 2, c = 1 and d = 1. In that case you would write
5√(x+2)+1 -1*x = 0

Then solve that equation for x (square everything and use the quadratic formula)
After you have your 2 x values, plug them both back into
5√(x+2)+1 -1*x = 0
If the value for x makes the left side not equal 0, it is an erroneous solution and you can throw it away. If it does make it equal 0, it is a good solution.

To find a radical expression with 2 solutions, first change the constants in the calculator so that the curve intersects in 2 places, then repeat the stuff from above.