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Question Number 128166 by bramlexs22 last updated on 05/Jan/21

(√(x−2)) + (√(x+3)) +(√(4x+1)) = 10 for x∈R

x−2+x+3+4x+1=10forx∈R

Answered by liberty last updated on 05/Jan/21

(√(4x+1)) = [ −(√(x+3))−(√(x−2))+10 ] 4x+1 = 2(√(x−2 )) (√(x+3)) −20(√(x+3)) +2x −20(√(x−2)) +101 (√(x+3)) = ((x+10(√(x−2))−50)/( (√(x−2))−10)) x+3 = (x^2 /(x−20(√(x−2))+98)) + ((20x(√(x−2)))/(x−20(√(x−2)) +98)) −((1000(√(x−2)))/(x−20(√(x−2))+98)) +((2300)/(x−20(√(x−2))+98)) (√(x−2)) =((101x−2006)/(40x−940)) x−2 = (((101x−2006)^2 )/((40x−940)^2 )) x = −((399(√(401))−79001)/(3200)) ; x=((399(√(401)) +79001)/(3200)) ; x=6

4x+1=[−x+3−x−2+10]4x+1=2x−2x+3−20x+3+2x−20x−2+101x+3=x+10x−2−50x−2−10x+3=x2x−20x−2+98+20xx−2x−20x−2+98−1000x−2x−20x−2+98+2300x−20x−2+98x−2=101x−200640x−940x−2=(101x−2006)2(40x−940)2x=−399401−790013200;x=399401+790013200;x=6

Commented by MJS_new last updated on 05/Jan/21

obviously x=6 is the only real solution. by squaring you introduce false solutions, you have to check each single one inserting it in the original equation

obviouslyx=6istheonlyrealsolution.bysquaringyouintroducefalsesolutions,youhavetocheckeachsingleoneinsertingitintheoriginalequation

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