How do you solve #1/2sqrt(2x-5)-1/2sqrt(3x+4)=-1#?

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Answer 1 · 2017-09-30T14:10:20

#x=15 or x=7#

#1/2sqrt(2x-5)-1/2sqrt(3x+4) =-1" "larr# multiply through by #2#

#sqrt(2x-5)-sqrt(3x+4) =-2#

#(sqrt(2x-5)-sqrt(3x+4))^2 =(-2)^2" "larr# square both sides

#(2x-5) - 2sqrt(2x-5)xxsqrt(3x+4) +(3x+4) = 4#

#2x-5 - 2sqrt(2x-5)xxsqrt(3x+4) +3x+4 -4 = 0#

#5x-5 = 2sqrt(2x-5)xxsqrt(3x+4)#

#(5x-5)^2 = (2sqrt(2x-5)xxsqrt(3x+4))^2" "larr# square both sides

#25x^2 -50x+25 = 4(2x-5)(3x+4)#

#25x^2 -50x+25 = 4(6x^2+8x-15x-20)#

#25x^2 -50x+25 = 24x^2+32x-60x-80#

#x^2 -22x +105=0" "larr# factorise

#(x-15)(x-7)=0#

#x =15 or x=7#