Check my answers.. I know one of them is incorrect..

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Check my answers.. I know one of them is incorrect..

Mathematics
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Solve. 4[4- 3/2x] +9=41 [] is absolute value I got x=8 and -8/3
Find all the real solutions. 6x^2/3 -41x ^1/3 -7=0 I got x= -1/216 and 343
If the first equation is \[\Large 4\left|4-\frac{3}{2}x\right|+9 = 41\] I'm getting x = 8 or x = -8/3 also. So both of your answers are correct (assuming I wrote the problem above correctly)

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3. Solve the polynomial and zero-product. 98x-49= 2x^3 -x^2 I got x=1/2 and -7 and 7
Yes thats the correct problem
4. Solve rational exponent. (x-6) ^3/2 =125 I got x=31 5. Solve radical equation. Square root 2x+11 =x+4 - the square root has 11 under it. I got -3 +- 3i square root 2
`Find all the real solutions.` `6x^2/3 -41x ^1/3 -7=0` `I got x= -1/216 and 343` I'm getting the same solutions as well
Okay
`3. Solve the polynomial and zero-product.` `98x-49= 2x^3 -x^2` `I got x=1/2 and -7 and 7` all three solutions are correct
#4 is correct but #5 is not correct
Okay... Hmmm where did I mess up :/
x^2 +6x+27=0 Is that part correct??
#5 is this problem right? \[\Large \sqrt{2x+11} =x+4\]
Yes
Square both sides to get... \[\Large \sqrt{2x+11} =x+4\] \[\Large \left(\sqrt{2x+11}\right)^2 =(x+4)^2\] \[\Large 2x+11 =x^2+8x+16\]
Yep I have that
I see where i messed up I added 11 instead of -11
Then get everything to one side \[\Large 2x+11 =x^2+8x+16\] \[\Large 2x+11{\color{red}{-11}} =x^2+8x+16{\color{red}{-11}}\] \[\Large 2x=x^2+8x+5\] \[\Large 2x{\color{red}{-2x}}=x^2+8x+5{\color{red}{-2x}}\] \[\Large 0=x^2+6x+5\] \[\Large x^2+6x+5=0\]
Yeah have that now
ok great
once you have your possible solutions, check them back in the original equation
x=-1 and x=-5
check each of those possible solutions back in the original equation
only -1 is true
Thank you!
yes -5 is extraneous

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