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Question Number 100158 by Algoritm last updated on 25/Jun/20

Answered by MJS last updated on 25/Jun/20

u=(√(x+3)) ⇔ x=u^2 −3∧u≥0 (u+4))(√(3u^2 −11))−3)=3u^2 −20 (u+4)(√(3u^2 −11))=3u^2 +3u−8 u^4 −u^3 −((38)/3)u^2 +((20)/3)u+40=0 trying factors of 40 ⇒ u_1 =3 ⇒ x_1 =6 u^3 +2u^2 −((20)/3)u−((40)/3)=0 trying ⇒ u_2 =−2 not possible because u≥0 u^2 −((20)/3)=0 ⇒ u_(3, 4) =±((2(√(15)))/3) ⇒ x_2 =((11)/3)

u=x+3⇔x=u2−3∧u⩾0(u+4))3u2−11−3)=3u2−20(u+4)3u2−11=3u2+3u−8u4−u3−383u2+203u+40=0tryingfactorsof40⇒u1=3⇒x1=6u3+2u2−203u−403=0trying⇒u2=−2notpossiblebecauseu⩾0u2−203=0⇒u3,4=±2153⇒x2=113

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