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Question Number 208652 by efronzo1 last updated on 20/Jun/24
Answered by Berbere last updated on 20/Jun/24
a,bsolutionof−3x3+2x=cS1=∫0ac−(2x−3x3)=∫ab(2x−3x3−c)dx=ca−a2−34a4=b2−a2−34(b4−a4)−c(b−a)b(b−34b3−c)=0c=b−34b3⇒2b−3b3=b−34b3⇒b−94b3=0⇒b2=49⇒b=23c=23−34(23)3=23−29=49−3x3+2x−49x3−23x+427=(x−23)(x2+23x−29)=0129;−23+1292=3−13=aa=3−13;b=23;c=49
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