How do you solve #sqrt(3x^2-11)=x+1# and identify any restrictions?

1 Answer
Mar 29, 2018

Answer:

#x=3#

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"#