Some Help need to solve equations for x

toshiba70

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1.27: Solve the following equations for \(\displaystyle x.\)

a) \(\displaystyle \sqrt{\strut{3x\, -\, 9\,}}\, -\, \sqrt{\strut{2x\, -\, 5\,}}\, =\, 1\)

b) \(\displaystyle \displaystyle{ 9^{\,x^2\, -\, 1}\, -\, 36\, \dot\, \,3^{\,x^2\, -\, 3}\, +\, 3\, =\, 0}\)

c) \(\displaystyle 2\, \sqrt[3]{\strut{x\, +\, 2\,} }\, +\, 3\, =\, \sqrt{\strut{15\, +\, 3x\, }}\)

Ex.1.27(C) Actually i think it is a formula (a+b)^3
 
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1.27: Solve the following equations for \(\displaystyle x.\)

a) \(\displaystyle \sqrt{\strut{3x\, -\, 9\,}}\, -\, \sqrt{\strut{2x\, -\, 5\,}}\, =\, 1\)

c) \(\displaystyle 2\, \sqrt[3]{\strut{x\, +\, 2\,} }\, +\, 3\, =\, \sqrt{\strut{15\, +\, 3x\, }}\)
To learn how to solve radical equations, please try here.

b) \(\displaystyle \displaystyle{ 9^{\,x^2\, -\, 1}\, -\, 36\, \dot\, \,3^{\,x^2\, -\, 3}\, +\, 3\, =\, 0}\)
To learn how to solve exponential equations, please try here.

Ex.1.27(C) Actually i think it is a formula (a+b)^3
I'm sorry, but I don't know what this means...?

Once you have studied at least two lessons from each link, please attempt the exercises. If you get stuck, you can then reply with a clear listing of your efforts so far. Thank you! ;)
 
1.27: Solve the following equations for \(\displaystyle x.\)

a) \(\displaystyle \sqrt{\strut{3x\, -\, 9\,}}\, -\, \sqrt{\strut{2x\, -\, 5\,}}\, =\, 1\)
To get rid of a square root, square! I would add \(\displaystyle \sqrt{\strut{2x- 5}}\) to both sides in order to hav a square root on both sides.
But you will still need to square twice to get rid of both square roots.

b) \(\displaystyle \displaystyle{ 9^{\,x^2\, -\, 1}\, -\, 36\, \dot\, \,3^{\,x^2\, -\, 3}\, +\, 3\, =\, 0}\)
\(\displaystyle 9= 3^2\) so this is \(\displaystyle (3^2)^{x^2- 1}- 36 (3^{x^2- 3}+ 3= 0\)
\(\displaystyle (3^2)^{x^2- 1}= 3^{2x^2- 2}= (3^x)^2(1/9)\) and
\(\displaystyle 3^{x^2- 3}= (3^x)^2(1/27)\)

Let \(\displaystyle y= 3^x\) and this will be a quadratic equation for y.

c) \(\displaystyle 2\, \sqrt[3]{\strut{x\, +\, 2\,} }\, +\, 3\, =\, \sqrt{\strut{15\, +\, 3x\, }}\)

Ex.1.27(C) Actually i think it is a formula (a+b)^3
Again, you have to take powers to get rid of the radicals. But now you will need to both square and take the third power.

(Be sure to check your answers in the original equation. Squaring or taking general powers may introduce "spurious" solutions that satisfy the new equation but not the old one.)
 
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Thank you Guys i solve it View attachment 4577

toshiba70,

your work starts off with your first line/step being incorrect. After you square each side,
if done correctly, it would look like the following:


\(\displaystyle (2\sqrt[3]{x + 2} \ + \ 3)(2\sqrt[3]{x + 2} \ + \ 3) \ = \ 15 \ + \ 3x \ \implies\)

\(\displaystyle 4(\sqrt[3]{x + 2})^2 \ + \ 12\sqrt[3]{x + 2} \ + \ 9 \ = \ 15 \ + \ 3x \ \)



You have:

\(\displaystyle 4\sqrt{x + 2} \ + \ 12\sqrt[3]{x + 2} \ + \ 9 \ = \ 15 \ + \ 3x\)


Then you may have fudged your work to come up with a correct equation later on.


Also, your answers have to be x-values, not the z-values that you boxed in. You
must substitute back.

.
.
 
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