Help please

DanieldeLucena said:
[attachment=0:3sqfz39i]1.JPG[/attachment:3sqfz39i]

One way to do it would be to:

substitute x = a[sup:3sqfz39i]1/2[/sup:3sqfz39i]

solve the quadratic equation and solve for 'x' and solve for 'a'

Then evaluate a + a[sup:3sqfz39i]-1[/sup:3sqfz39i]

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
DanieldeLucena said:
[attachment=0:23tcznuf]1.JPG[/attachment:23tcznuf]

Hmmmm....what would happen if you SQUARED both sides of that equation?

Why don't you try that, and see if something useful emerges.

If you're still stuck, post again, showing us your efforts.
 
[attachment=0:2dozopet]1.JPG[/attachment:2dozopet]

x=a^1/2

3=a^1/2 ???? and 1/3=a^1/2 ????

I don't know how to continue
 
I was thinking like this, but I don't know how this way too
[attachment=0:6fue9vpn]1.JPG[/attachment:6fue9vpn]
 
Mrspi said:
DanieldeLucena said:
[attachment=0:3p0139c1]1.JPG[/attachment:3p0139c1]

Hmmmm....what would happen if you SQUARED both sides of that equation?

Why don't you try that, and see if something useful emerges.

If you're still stuck, post again, showing us your efforts.

I was referring to the original equation:

(a[sup:3p0139c1]1/2[/sup:3p0139c1] + a[sup:3p0139c1]-1/2[/sup:3p0139c1])[sup:3p0139c1]2[/sup:3p0139c1] = (10/3)[sup:3p0139c1]2[/sup:3p0139c1]

On the left side, we'll need the pattern for squaring a binomial.
The pattern for squaring a binomial is this: (m + n)[sup:3p0139c1]2[/sup:3p0139c1] = m[sup:3p0139c1]2[/sup:3p0139c1] + 2 m n + n[sup:3p0139c1]2[/sup:3p0139c1]

replace "m" with a[sup:3p0139c1]1/2[/sup:3p0139c1]
replace "n" with a[sup:3p0139c1]-1/2[/sup:3p0139c1]

(a[sup:3p0139c1]1/2[/sup:3p0139c1] + a[sup:3p0139c1]-1/2[/sup:3p0139c1])[sup:3p0139c1]2[/sup:3p0139c1] = (a[sup:3p0139c1]1/2[/sup:3p0139c1])[sup:3p0139c1]2[/sup:3p0139c1] + 2*a[sup:3p0139c1]1/2[/sup:3p0139c1]*a[sup:3p0139c1]-1/2[/sup:3p0139c1] + (a[sup:3p0139c1]-1/2[/sup:3p0139c1])[sup:3p0139c1]2[/sup:3p0139c1]

Now you need to apply some of the rules for exponents....when you raise a power to a power, you MULTIPLY the exponents. When you multiply two powers of the same base, you ADD the exponents.

a[sup:3p0139c1](1/2)*2[/sup:3p0139c1] + 2*a[sup:3p0139c1]1/2 + (-1/2)[/sup:3p0139c1] + a[sup:3p0139c1](-1/2)*2[/sup:3p0139c1] = 100/9

See what you can do with that now....
 
I don't think the book's answer is wrong.....


a[sup:1rndw1ss](1/2)*2[/sup:1rndw1ss] + 2*a[sup:1rndw1ss]1/2 + (-1/2)[/sup:1rndw1ss] + a[sup:1rndw1ss](-1/2)*2[/sup:1rndw1ss] = 100/9


Apply the rules of exponents:

a + 2*a[sup:1rndw1ss]0[/sup:1rndw1ss] + a[sup:1rndw1ss]-1[/sup:1rndw1ss] = 100/9

But what is a[sup:1rndw1ss]0[/sup:1rndw1ss]? As long as "a" is a non-zero number, a[sup:1rndw1ss]0[/sup:1rndw1ss] = 1.

a + 2*1 + a[sup:1rndw1ss]-1[/sup:1rndw1ss] = 100/9

(a + a[sup:1rndw1ss]-1[/sup:1rndw1ss]) + 2 = 100/9

Subtract 2 from both sides. Combine terms on the right side....remember that you'll need to have fractions with the same denominator before you can add or subtract.
 
DanieldeLucena said:
[attachment=0:26s0tgdk]1.JPG[/attachment:26s0tgdk]

x=a^1/2

3=a^1/2 ???? and 1/3=a^1/2 ????

I don't know how to continue

a[sup:26s0tgdk]1/2[/sup:26s0tgdk] = 3

a = 9

a + a[sup:26s0tgdk]-1[/sup:26s0tgdk] = 9 + 1/9 = 82/9

However, Mrspi's solution is more elegant.
 
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