Choose any four (4) of the five radical equations below. Solve each equation, showing all steps
leading to your answer. Then check your answer to determine if any of the solutions are
extraneous
1. √5x - x = 0
2. √(2x + 1) + 1 = x
3. √(3x - 2) + 3 = 4x
can someone please help me solve it ?

- anonymous

- schrodinger

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- triciaal

steps
isolate the radicand to one side of the equation. remember whatever you do to one side also do to the other
square each side to get rid of the radicand and maintain the balance
isolate the terms with the variable to one side
simplify if possible
divide by the coefficient of the variable
take a minute to review then I'll walk you through one after which you can try the others on your own and I will verify if you show your work and ask me to

- triciaal

|dw:1441428621396:dw|

- triciaal

problem 3
the radicand is labeled
to isolate by subtracting 3
square each side and simplify.
this gives a quadratic we can solve by factoring/completing the square/ graphing/matrices or using the formula.
minor note to the steps I wrote if it was a linear equation divide by the coefficient of the variable if like in this case you have a quadratic solve using one of the 5 methods

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## More answers

- triciaal

are you with me?

- triciaal

do you know how to solve quadratic equations?

- anonymous

do we factor it ?

- triciaal

yes one of the methods

- anonymous

wait , i am bit lost .
when u square (4x - 3)^2 , isn't the answer 16x^2 - 12x + 9 ?

- triciaal

good catch thanks
|dw:1441430085255:dw|

- triciaal

|dw:1441430227757:dw|

- anonymous

so now we factor it ?

- anonymous

how do we factor it ?

- triciaal

##### 1 Attachment

- triciaal

I am going to use the formula

- triciaal

|dw:1441430939433:dw|

- triciaal

|dw:1441430974635:dw|

- triciaal

reply with above and identify your a, b, c

- anonymous

ok
so a = 16 , b = 27 , c = 11 right ?

- anonymous

then using the above formula we will have : |dw:1441431330504:dw|

- triciaal

|dw:1441431722636:dw|

- anonymous

|dw:1441431470048:dw|

- triciaal

|dw:1441431767526:dw|

- anonymous

is my answer correct ?

- triciaal

|dw:1441431858754:dw|

- triciaal

it can't be because you have the wrong C

- triciaal

|dw:1441432220054:dw|

- anonymous

isn't 2 + 9 = 11 ?
since 3x - 2 = 16x^2 - 24x + 9
16x^2 - 24x - 3x + 9 + 2
= 16x^2 - 27x + 11 ?

- triciaal

think we have another error the graph shows 2.027

- triciaal

think I am too tired to be on here but happy you grasp something sorry need to be off right now

- anonymous

ok so i think i am understanding the question number 3. thank you so much for ur help

- anonymous

i was hoping u would help in question number 1 or 2 ? but if ur busy then its ok

- Jhannybean

For question 1:
rewrite your function: \(\sqrt{5}x-x=0\)
\[(\sqrt{5}-1)x=0\]\[\boxed{x= \frac{0}{\sqrt{5}-1} = 0}\]

- Jhannybean

Is it \(\sqrt{5}x\) or \(\sqrt{5x}\)?

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