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

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

$x =15 or x=7$

How do you solve 1/2sqrt(2x-5)-1/2sqrt(3x+4)=-11/2sqrt(2x-5)-1/2sqrt(3x+4)=-1?