anonymous
  • anonymous
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 ?
Pre-Algebra
schrodinger
  • schrodinger
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triciaal
  • 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
  • triciaal
|dw:1441428621396:dw|
triciaal
  • 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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triciaal
  • triciaal
are you with me?
triciaal
  • triciaal
do you know how to solve quadratic equations?
anonymous
  • anonymous
do we factor it ?
triciaal
  • triciaal
yes one of the methods
anonymous
  • anonymous
wait , i am bit lost . when u square (4x - 3)^2 , isn't the answer 16x^2 - 12x + 9 ?
triciaal
  • triciaal
good catch thanks |dw:1441430085255:dw|
triciaal
  • triciaal
|dw:1441430227757:dw|
anonymous
  • anonymous
so now we factor it ?
anonymous
  • anonymous
how do we factor it ?
triciaal
  • triciaal
1 Attachment
triciaal
  • triciaal
I am going to use the formula
triciaal
  • triciaal
|dw:1441430939433:dw|
triciaal
  • triciaal
|dw:1441430974635:dw|
triciaal
  • triciaal
reply with above and identify your a, b, c
anonymous
  • anonymous
ok so a = 16 , b = 27 , c = 11 right ?
anonymous
  • anonymous
then using the above formula we will have : |dw:1441431330504:dw|
triciaal
  • triciaal
|dw:1441431722636:dw|
anonymous
  • anonymous
|dw:1441431470048:dw|
triciaal
  • triciaal
|dw:1441431767526:dw|
anonymous
  • anonymous
is my answer correct ?
triciaal
  • triciaal
|dw:1441431858754:dw|
triciaal
  • triciaal
it can't be because you have the wrong C
triciaal
  • triciaal
|dw:1441432220054:dw|
anonymous
  • anonymous
isn't 2 + 9 = 11 ? since 3x - 2 = 16x^2 - 24x + 9 16x^2 - 24x - 3x + 9 + 2 = 16x^2 - 27x + 11 ?
triciaal
  • triciaal
think we have another error the graph shows 2.027
triciaal
  • triciaal
think I am too tired to be on here but happy you grasp something sorry need to be off right now
anonymous
  • anonymous
ok so i think i am understanding the question number 3. thank you so much for ur help
anonymous
  • anonymous
i was hoping u would help in question number 1 or 2 ? but if ur busy then its ok
Jhannybean
  • 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
  • Jhannybean
Is it \(\sqrt{5}x\) or \(\sqrt{5x}\)?

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