How do you solve #sqrt(13-4x^2)=2-x# and identify any restrictions?

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Answer 1 · 2017-08-15T11:36:02

Solution: # x = -1 or x =9/5#

Restriction : # 13 - 4x^2 >=0 or 13 >= 4x^2 #

#13/4 >= x^2 or x^2 <= 13/4 or x <= +- sqrt13/2#

#x <= 1.8027756 or x >= -1.8027756#

# -1.8028 <= x <= 1.8028 or [-1.8028 ,1.8028]#

# sqrt(13-4x^2) =2-x # Squaring both sides we get,

#(13-4x^2) =(2-x)^2 or 13-4x^2 =4-4x+x^2 # or

#4x^2+x^2 -4x +4 -13=0 or 5x^2 - 4x -9 =0# or

# 5x^2 +5x - 9x -9 =0 or 5x(x+1) -9(x+1) =0# or

# (x+1)(5x-9)=0 # . Either #x+1 =0 :. x =-1 # or

#5x-9=0 :. 5x=9 or x =9/5# . Solution: # x = -1 or x =9/5#

Check:

When #x= -1 ; sqrt(13-4x^2) =sqrt(13-4(-1)^2)=sqrt 9 =3#

#2-x = 2 - (-1) =3 :. LHS =RHS # (verified).

When #x= 9/5 ; sqrt(13-4x^2) =sqrt(13-4*81/25)=sqrt 0.04 =0.2#

#2-x = 2 - 9/5 =2-1.8 = 0.2 :. LHS =RHS # (verified).

So both # x = -1 and x =9/5# hold good. [Ans]