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# Which of the following is an extraneous solution of mc016-1.jpg? x = –6 x = –1 x = 1 x = 6

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[Incomplete question. I found the equation in other source:√(-3X-2)=X+2] are we Going to find x:√(-3X-2)=X+2-3X-2=(X+2)2-3X-2=x2+2*2*x+22-3X-2=x2+4x+4-3X-2-x2-4x-4=0-x2-7x-6=0So now you must use the formula of Bhaskara. Unless your calculator does by itself (some calculators).ax2+bx+c=0 x=(-b+√(b2-4ac))/(2a)=-1b=-7c=-6x=-1 x=-6Let, verify -1√(-3*(-1)-2)=-1+2√(3-2)=1√1=11=1 TRUE -> So -1 is a solution, not a stranger solutionLet check x=-6:√(-3*(-6)-2)=-6+2√(18-2)=-4√16=-44=-4 FALSE –> So -6 is coming out of the equation, but not verify. So -6 is an extraneous solution. This is related to the fact that square roots have only 1 solution, while the squares are 2 solutions (remember (-1)2 is 1, and 12 is 1 as well, due to the rules of the signs for the multiplication).