My Math Forum Surds(need help)

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 June 4th, 2011, 01:33 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Surds(need help) 1.Find the square root of surdic quantities 73 +40?3 =( ?x + ?y)^2 73+40?3 = x+y + 2?xy x+y=73 y=73-x ..........(1) 2?xy=40?3 ?xy=20?3 xy=1200 ........(2) substitute (1) into (2) x(73-x)=1200 73x^2 - x^2 =1200 x^2-73x+1200=0 then i m stuck at here,i have problem in factorize it 2.Show that ?2x-3 - ?x+2 =3?x-1 does not have any real root. I know how to solve but i dunno how to show it does not have any real root. 3. Show that ?10+2?2 is less than 6 . This is the provided solution (?10+2?2)^2 = 10+ 2^2(2)+2(2)?10x2 = 18 + 4?20 = 18 + ?16x20 = 18+ ?320 (?10 + 2?2)^2 < 18 + ?324 ( i don understand how ?320 become ?324 ) < 18+18 < 36 ?10 + 2?2 < 6 Thank you .
 June 4th, 2011, 01:51 AM #2 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 3. ?10 + 2?2 < 6 is trivial. Square both sides and we get 18 + 8?5 < 36, i.e. ?5 < 9/4 = ?(81/16). ?5 = ?(80/16) < ?(81/16), hence ?5 < 9/4 is correct, i.e. 18 + 8?5 < 36 is correct. Therefore ?10 + 2?2 < 6 is correct.
 June 4th, 2011, 03:44 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Surds(need help) 1.) $73+40\sqrt{3}=$$\sqrt{x}+\sqrt{y}$$^2$ $25+2\sqrt{25}\sqrt{48}+48=$$\sqrt{x}+\sqrt{y}$$^2$ $$$\sqrt{25}+\sqrt{48}$$^2=$$\sqrt{x}+\sqrt{y}$$^2$ x = 25, y = 48 or x = 48, y = 25 2.) Use parentheses to clearly show what the radicands are. I will ASSUME: $\sqrt{2x-3}-\sqrt{x+2}=3\sqrt{x-1}$ $2x-3-2\sqrt{(2x-3)(x+2)}+x+1=9(x-1)$ $3x-2-2\sqrt{(2x-3)(x+2)}=9x-9$ $-(6x+11)=2\sqrt{(2x-3)(x+2)}$ $36x^2+132x+121=4$$2x^2+x-6$$=8x^2+4x-24$ $28x^2+128x+145=0$ $(2x+5)(14x+29)=0$ $x=-\frac{5}{2},-\frac{29}{14}$ Neither of these roots are in the domain of the original equation.
June 4th, 2011, 06:49 AM   #4
Math Team

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Re: Surds(need help)

Hello, hoyy1kolko!

Quote:
 $\text{1. Find the square root of surdic quantities: }\;73\,+\,40\sqrt{3} \:=\\sqrt{x}\,+\,\sqrt{y})^2" />

$\text{We have: }\;73\,+\,40\sqrt{3} \:=\\,+\,2\sqrt{xy}\,+\,y" />

$\text{Hence: }\;\begin{Bmatrix}x\,+\,y\;=\;73 \;\; [1] \\ 2\sqrt{xy} \:=\:40\sqrt{3} &\;\;\Rightarrow\;\;=&xy \:=\:1200 &\;\;\Rightarrow\;\;=&y \:=\:\frac{1200}{x} &[2] \end{Bmatrix}=$

$\text{Substitute [2] into [1]: }\;x\,+\,\frac{1200}{x} \:=\:73 \;\;\;\Rightarrow\;\;\;x^2\,-\,73x\,+\,1200 \:=\:0$

$\text{Factor: }\;(x\,-\,25)(x\,-\,4 \:=\:0 \;\;\;\Rightarrow\;\;\;x \:=\:25,\:48 \;\;\;\Rightarrow\;\;\;y \:=\:48,\:25" />

$\text{Therefore: }\:\sqrt{x}\,+\,\sqrt{y} \;=\;\sqrt{25}\,+\,\sqrt{48} \;=\;5\,+\,4\sqrt{3}$

 June 4th, 2011, 06:58 AM #5 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Surds(need help) Thank you very much,i still don understand the solution from johnny for question 3 ,i need more help. What is mean by trivial? and isn't he use comparison to solve the question?
 June 4th, 2011, 02:59 PM #6 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 What I mean by trivial is ?10 + 2?2 ?< 6, in another words we don't know if the inequality is true. Afterwards I did some simple mathematics to see that if it's true and then arrived at the correct conclusion.
 June 5th, 2011, 05:16 AM #7 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Surds(need help) Thank you very much!!!!!

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