Issues simplifying radical expressions, need help please!

[jayson]

I thank those who help me ahead of time. There must be something I am missing but what I am going to do is show you an example of my problems
so you can better understand my situation. So far I am only doing odd exercises only. I can't figure out 19, and 21, and 23. If you guys can help
me out I would appreciate it. I must be missing something.

 

[srmichael]

Show us what you have done so far so we can see where you are stuck.

 

[jayson]

Show us what you have done so far so we can see where you are stuck.

In 19 I don't know where to start so I don't have much to show. What I did try to do is move the a to left and the 5 to the right. That is where I got stuck.

 

[jayson]

No, move the 5 to the right:
SQRT(5 + a^2) = 5 - a
Now square both sides.
If you can't do that, see your teacher.

ok that is where I got lost the first time


5+a^2 = 25 - a^2

They are already squared and I got stuck there. Let me know if you can help.

 

[soroban]

Hello, jayson!

In 19 I don't know where to start.

That's just great! .You can't do the first problem.
This means you've never heard of Radical Equations ... ever!
So why were you assigned two dozen such problems?

Okay, here is your missing lesson.

Isolate the radical.
. . (Get the square root "alone" on one side of the equation.)
Square both sides of the equation ... carefully!
Solve the resulting equation.
Check all answers ... some may be extraneous.


\(\displaystyle (23)\;\;3 - \sqrt{7x-3} \:=\:2x\)

\(\displaystyle \text{We have: }\qquad\;\;\; 3 - 2x \;=\;\sqrt{7x-3}\)

\(\displaystyle \text{Square: }\qquad\;\;(3-2x)^2 \;=\;\left(\sqrt{7x-3}\right)^2\)

. . . . . . . .\(\displaystyle 9 - 12x + 4x^2 \;=\;7x-3\)

. . . . . . .\(\displaystyle 4x^2 - 19x + 12 \;=\;0\)

\(\displaystyle \text{Factor: }\;(4x-3)(x-4) \;=\;0\)

\(\displaystyle \text{Hence: }\:\begin{Bmatrix}4x-3 \:=\:0 & \Rightarrow & x \:=\:\frac{3}{4} \\ x-4\:=\:0 & \Rightarrow & x \:=\:4 \end{Bmatrix}\)


Check. \(\displaystyle x = \frac{3}{4}\)

\(\displaystyle \text{Does }\,3 - \sqrt{7(\frac{3}{4})-3} \;=\;2(\frac{3}{4})\:?\)

. . \(\displaystyle 3 - \sqrt{\frac{21}{4} - 3} \;\;=\;\; 3 - \sqrt{\frac{9}{4}} \;\;=\;\;3 - \frac{3}{2} \;\;=\;\;\frac{3}{2}\) .Yes!

Hence, \(\displaystyle x = \frac{3}{4}\) is a solution.


Check..\(\displaystyle x = 4\)

\(\displaystyle \text{Does }\,3 - \sqrt{7(4) - 3} \;=\;2(4)\,?\)

. . \(\displaystyle 3 - \sqrt{25} \;\;=\;\;3-5 \;\;=\;\;-2\) .No!

Hence, \(\displaystyle x = 4\) is an extraneous solution.


Therefore, the only solution is. \(\displaystyle x = \frac{3}{4}\)

 

[jayson]

Hello, jayson!


That's just great! .You can't do the first problem.
This means you've never heard of Radical Equations ... ever!
So why were you assigned two dozen such problems?

Okay, here is your missing lesson.

Isolate the radical..(Get the square root "alone" on one side of the equation.)
Square both sides of the equation ... carefully!
Solve the resulting equation.
Check all answers ... some may be extraneous.


\(\displaystyle (23)\;\;3 - \sqrt{7x-3} \:=\:2x\)

\(\displaystyle \text{We have: }\qquad\;\;\; 3 - 2x \;=\;\sqrt{7x-3}\)

\(\displaystyle \text{Square: }\qquad\;\;(3-2x)^2 \;=\;\left(\sqrt{7x-3}\right)^2\)

. . . . . . . .\(\displaystyle 9 - 12x + 4x^2 \;=\;7x-3\)

. . . . . . .\(\displaystyle 4x^2 - 19x + 12 \;=\;0\)

\(\displaystyle \text{Factor: }\;(4x-3)(x-4) \;=\;0\)

\(\displaystyle \text{Hence: }\:\begin{Bmatrix}4x-3 \:=\:0 & \Rightarrow & x \:=\:\frac{3}{4} \\ x-4\:=\:0 & \Rightarrow & x \:=\:4 \end{Bmatrix}\)


Check. \(\displaystyle x = \frac{3}{4}\)

\(\displaystyle \text{Does }\,3 - \sqrt{7(\frac{3}{4})-3} \;=\;2(\frac{3}{4})\:?\)

. . \(\displaystyle 3 - \sqrt{\frac{21}{4} - 3} \;=\; 3 - \sqrt{\frac{9}{4}} \;=\;3 - \frac{3}{2} \;=\;\frac{3}{2}\) .Yes!

Hence, \(\displaystyle x = \frac{3}{4}\) is a solution.


Check..\(\displaystyle x = 4\)

\(\displaystyle \text{Does }\,3 - \sqrt{7(4) - 3} \;=\;2(4)\,?\)

. . \(\displaystyle 3 - \sqrt{25} \;=\;3-5 \;=\;-2\) .No!

Hence, \(\displaystyle x = 4\) is an extraneous solution.


Therefore, the only solution is. \(\displaystyle x = \frac{3}{4}\)

Thanks for the help, this may be easy for you but if you have noticed, I did exercises from 1 to 19 at this point. This means I completed a majority of the exercises and they are getting progressively harder.
I think you would have noticed that the exercises that I highlighted are actually the last 10 it would be nice of you not to assume that I haven't done anything to help myself. So number 19 would not be
the first problem it would be the 19th problem.

 

[srmichael]

(5 - a)^2 is NOT = 25 - a^2 ; it equals 25 - 10a + a^2
CAPISH?

Top 5 Classic Math Mistakes:

1) \(\displaystyle (a+b)^2=a^2+b^2\)
2) \(\displaystyle -x^2=x^2\)
3) If \(\displaystyle x^2=4\) then \(\displaystyle x=2\)
4) \(\displaystyle a^{-m}=-a^m\)
5) I'm confused...I'll choose E) None of the above

:shock:

 

[jayson]

Ok guys for 25 here is my work ! Let me know if its right!

 

[jayson]

 

[jayson]

#25 is correct...good stuff!
Can't tell about the others...too faint !

By the way, you can check if your answer is correct
by substituting it in the original equation;
with #25: is SQRT[4(48) / 3] -2 = 6 ?

I verified and they are correct. Used some of the tools you posted on the 23 I think to solve the other ones.