Calculate: (x+2(x-1)^(1/2))^(1/2)+(x-2(x-1)^(1/2))^(1/2) for x>2 and for 1>=x>=2
\[\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\]
square the binomial: \[ a^2=(x+\cancel{2\sqrt{x-1}})+2\sqrt{(x+2\sqrt{x-1})(x-2\sqrt{x-1})}+(x-\cancel{2\sqrt{x-1}})\\ a^2=2x+2\sqrt{x^2-4(x-1)}\\ a^2=2x+2\sqrt{x^2-4x+4}\\ a^2=2x+2\sqrt{(x-2)^2}\\ a^2=2x+2(x-2)\\ a^2=2x+2x-4\\ a^2=4(x-1)\\ \Large\boxed{a=\pm2\sqrt{x-1}} \] but we choose only the positive value since the given square-root problem is always positive.
formulae use: \[ (a+b)^2=a^2+2ab+b^2\\ (a+b)(a-b)=a^2-b^2 \]
did you follow?
yep the only problem is that this is actually a problem of absolute value of a real number
right, |a|
when u take the square off the root u should absolute it, because this is like for 1>=x>=2 a=2 and for x>2
explain that the only condition needed here would be \(x\ge1\) your other condition does not make sense. \(1\ge x\ge2\) means that x is less than or equal to 1 AND greater than or equal to 2 -> an illogical statement.
if x<1, the radicals become undefined.
its the oposite 2>=x>=1
the answer to this problem is for 2>=x>=1 the expression is 2 and for x>2 the expression becomes 2(x-1)^(1/2)
right, that is looking at the domain \[ a) \;x-1\ge0\implies x\ge 1\\ b)\; x+2\sqrt{x-1}\ge0\qquad x-2\sqrt{x-1}\ge0\\ \implies x\ge 2\sqrt{x-1}\\ \implies x^2\ge 4(x-1)\\ \implies x^2-4x+4\ge0\\ \implies |x-2|\ge0\\ \implies x-2\ge 0\\ \implies x\ge 2 \] resolving, \[ a^2=(x+\cancel{2\sqrt{x-1}})+2\sqrt{(x+2\sqrt{x-1})(x-2\sqrt{x-1})}+(x-\cancel{2\sqrt{x-1}})\\ a^2=2x+2\sqrt{x^2-4(x-1)}\\ a^2=2x+2\sqrt{x^2-4x+4}\\ a^2=2x+2\sqrt{(x-2)^2}\\ a^2=2x+2\color{red}{|x-2|}\\ a^2=2x+2x-4\quad(\forall x\ge2)\qquad a^2=2x-2x+4\quad (\forall 1\le x\le2)\\ a^2=\cases{4(x-1)\qquad \forall \;x\ge2\\ 4\qquad \forall\;1\le x\le2}\\ \Large\boxed{a=\cases{2\sqrt{x-1}\quad\forall\; x>2\\2\qquad\qquad \forall\;1\le x\le2}} \]
Thx a lot man, sorry for the confusion!!!
yw
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