How do you solve #sqrt(3x^2-11)=x+1# and identify any restrictions?
1 Answer
Mar 29, 2018
Explanation:
#color(blue)"square both sides"#
#rArr(sqrt(3x^2-11))^2=(x+1)^2#
#rArr3x^2-11=x^2+2x+1#
#rArr2x^2-2x-12=0larrcolor(blue)"in standard form"#
#rArr2(x^2-x-6)=0#
#rArr2(x-3)(x+2)=0#
#"equate each factor to zero and solve for x"#
#x-3=0rArrx=3#
#x+2=0rArrx=-2#
#color(blue)"As a check"# Substitute these values into the equation and if both sides are equal then they are the solutions.
#sqrt(27-11)=sqrt16=4" and "3+1=4#
#sqrt(12-11)=1" and "-2+1=-1#
#"the solution is "x=3#
#x=-2" is an extraneous solution"#
