Question 1

Why do we get extraneous solutions? At which point during the solution process causes this?

Accepted Answer

Imagine this: There's an equation x = 2.
Right now, x is only equal to two.
Square both sides, and x^2 = 4.
For some reason, if you want to take the square root of both sides, and you get x= +/- 2, because -2 squared is still equal to four.
But, according to the original equation, x is only equal to 2.
Therefore -2 is an extraneous solution, and squaring both sides of the equation creates them.

Question 2

In the last problem how does it equal 30x and not 15x to begin with in step 2?

Accepted Answer

Hey there friend!

So in Step 2 you get ((5+3x)^2 which is equivalent to (5+3x)*(5+3x). 
After you distribute you get this string of numbers: 25+15x+15x+9x^2
From there you simplify to: 9x^2+30x+25.

I hope this helps! Have a good day! :D

Question 3

how do u work out the the square root of n equal square root of 2n plus 6

Accepted Answer

sqrt(n) = sqrt(2n+6)
To undo a square root, you raise it to the 2nd power (square it). What you do on one side you must do on the other side. 
So sqrt(n)^2 = n while sqrt(2n+6)^2 = 2n+6
This will leave you with n=2n+6 to solve for n.

Question 4

(2x + 5)(x + 5) = 0 How did you get this?

Accepted Answer

Factor the quadratic using grouping.  See the videos at this link:  https://www.khanacademy.org/math/algebra/polynomial-factorization#factoring-quadratics-2

Question 5

Interesting, the videos didn't discuss (that I saw) there being NO solutions.

Accepted Answer

That is covered in a later video in this lesson.

Question 6

*Is there a specific way to solve for square roots or is it always just going to be trial and error?*

Accepted Answer

We can use a linear approximation to find a close estimate for the square root. √(x) ≈ (x + y) / (2 * √(y)) where y is a number that is "close to" x. Typically, you would choose y to be a perfect square to make the math easy.

Question 7

For practice question #1, why couldn't I square both sides- the squareroot of 2x-7 -3 and then the 4? Wouldn't that keep the equations equivalent?

Accepted Answer

You could, but it doesn't eliminate the radical.

You may be thinking that if you square the left side that it becomes 2x-7+9.  It doesn't.  (a+b)^2 never equals a^2+b^2.

Here's what happens if you square both sides immediately
[sqrt(2x-7)-3]^2 = 4^2
The left side is a binomial squared, so use FOIL or extended distribution to square that side.
2x-7 -3sqrt(2x-7) - 3sqrt(2x-7) + 9 = 16
2x+2-6sqrt(2x-7) = 16
You have just made the equation more complicated, and it still has a radical.

This is why the correct 1st step is to isolate the radical *before* squaring both sides.

Hope this helps.

Question 8

For root(7x^2 + 15x) - 2x = 5 + x
Got solutions as x = -5 and x = -(5/2)
While checking for extraneous Solutions - 
For x = -5 we get 
root(100) + 10 = 0
The solution mentioned on the website considers only positive value of root(100) i.e. +10. Shouldn't it also consider -10 ?

Accepted Answer

There is a difference between taking the square root of a number which is always positive (√100=10) and solving x^2=100 which gives both a positive and negative answer.  The first is finding a value on the square root function, the second is finding the x intercepts of an equation.  Does this make sense?
I looked on DESMOS and they did not cross as a check.  This means both answers are extraneous.

Algebra II (2018 edition)

Course: Algebra II (2018 edition) > Unit 5

Solving square-root equations

Introduction

Practice question 1: Isolating the radical term

Practice question 2: Two possible solutions

Practice question 3

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