Pre-Cal questions

[BonnieBrown011]

Need help solving these 2

16/16^1/3 Rationalize the denominator

2x^1/2-(2x+1)^1/2=1 solve the equation

 

[Denis]

We can't tell where you're stuck if you don't show your work.

 

[BonnieBrown011]

16/(16^1/3) . 16^1/3 / 16^1/3 = 16(16^1/3)/16= 16^1/3 answer given is 4(4^1/3)

2(2^1/2)-(2x=1)^1/2=1 square all
4x-2x+1=1

given answer is 4

 

[Deleted member 4993]

16/(16^1/3) . 16^1/3 / 16^1/3 = 16(16^2/3)/16= 16^2/3 = [(4)[sup:1gm1szj5]2[/sup:1gm1szj5]][sup:1gm1szj5]2/3[/sup:1gm1szj5] = 4[sup:1gm1szj5]4/3[/sup:1gm1szj5]

Now continue....
answer given is 4(4^1/3)

2(2^1/2)-(2x=1)^1/2=1 square all <<< This does not make any sense - why do you have "=" sign inside parenthesis? Please check your work and put corrected work
4x-2x+1=1

given answer is 4

 

[BonnieBrown011]

2(x^1/2) - (2x+1)^1/2=1 square all
4x-2x+1=1
2x+1=1

given answer is 4

 

[Deleted member 4993]

2(x^1/2) - (2x+1)^1/2=1 square all
4x-2x+1=1<<< This is wrong - remember (a - b)[sup:1v488cbl]2[/sup:1v488cbl] = a[sup:1v488cbl]2[/sup:1v488cbl] + b[sup:1v488cbl]2[/sup:1v488cbl] -2ab
2x+1=1

given answer is 4

 

[BonnieBrown011]

Could you elaborate a bit more? I do recall the equation shown, but am having difficulty relating it to the problem.

 

[Deleted member 4993]

2(x^1/2) - (2x+1)^1/2=1 square all
4x-2x+1=1<<< This is wrong - remember (a - b)[sup:2oqtov4f]2[/sup:2oqtov4f] = a[sup:2oqtov4f]2[/sup:2oqtov4f] + b[sup:2oqtov4f]2[/sup:2oqtov4f] -2ab
2x+1=1

given answer is 4

a = 2x[sup:2oqtov4f]1/2[/sup:2oqtov4f] ? a[sup:2oqtov4f]2[/sup:2oqtov4f] = 4x

b = (2x+1)[sup:2oqtov4f]1/2[/sup:2oqtov4f] ? b[sup:2oqtov4f]2[/sup:2oqtov4f] = 2x + 1

ab = 2x[sup:2oqtov4f]1/2[/sup:2oqtov4f] * (2x+1)[sup:2oqtov4f]1/2[/sup:2oqtov4f] = 2(2x[sup:2oqtov4f]2[/sup:2oqtov4f] + x)[sup:2oqtov4f]1/2[/sup:2oqtov4f]

[2x[sup:2oqtov4f]1/2[/sup:2oqtov4f] - (2x+1)[sup:2oqtov4f]1/2[/sup:2oqtov4f]][sup:2oqtov4f]2[/sup:2oqtov4f] = 4x + (2x+1) - 2*2(2x[sup:2oqtov4f]2[/sup:2oqtov4f] + x)[sup:2oqtov4f]1/2[/sup:2oqtov4f] .... and continue

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[masters]

Could you elaborate a bit more? I do recall the equation shown, but am having difficulty relating it to the problem.

Hi BonnieBrown011,

Let's convert to radicals.

\(\displaystyle 2\sqrt{x}-\sqrt{2x+1}=1\)

\(\displaystyle 2\sqrt{x}=1+\sqrt{2x+1}\)

\(\displaystyle \left(2\sqrt{x}\right)^2=\left(1+\sqrt{2x+1}\right)^2\)

\(\displaystyle 4x=1+2\sqrt{2x+1}+2x+1\)

\(\displaystyle 2x-2=2\sqrt{2x+1}\)

\(\displaystyle x-1=\sqrt{2x+1}\)

\(\displaystyle \left(x-1\right)^2=\left(\sqrt{2x+1}\right)^2\)

\(\displaystyle x^2-2x+1=2x+1\)

\(\displaystyle x^2-4x=0\)

\(\displaystyle x(x-4)=0\)

\(\displaystyle x=0 \:\r\:\:x=4\)

Upon checking, you can see that x = 0 is not a solution.