Solve the equation?

sqrt(x^2+4x-21)+sqrt(x^2-x-6)=sqrt(6x^2-5x-39)

1 Answer

x=3

Explanation:

sqrt(x^2+4x-21)+sqrt(x^2-x-6)=sqrt(6x^2-5x-39)

(sqrt(x^2+4x-21)+sqrt(x^2-x-6))^2=(sqrt(6x^2-5x-39))^2

x^2+4x-21+2*sqrt(x^2+4x-21)*sqrt(x^2-x-6)+x^2-x-6=6x^2-5x-39

2x^2+3x-27+2sqrt((x^2+4x-21)(x^2-x-6))=6x^2-5x-39

2sqrt((x^2+4x-21)(x^2-x-6))=4x^2-8x-12

sqrt((x^2+4x-21)(x^2-x-6))=2x^2-4x-6

(sqrt((x^2+4x-21)(x^2-x-6)))^2=(2x^2-4x-6)^2

(x^2+4x-21)(x^2-x-6)=(2)^2(x^2-2x-3)^2

(x+7)(x-3)(x-3)(x+2)=4((x-3)(x+1))^2

(x+7)(x-3)^2(x+2)=4(x-3)^2(x+1)^2

(x-3)^2((x+7)(x+2)-4(x+1)^2)=0

(x-3)^2(x^2+9x+14-4(x^2+2x+1))=0

(x-3)^2(x^2+9x+14-4x^2-8x-4)=0

(x-3)^2(-3x^2+x+10)=0

-(x-3)^2(3x^2-x-10)=0

-(x-3)^2(3x+5)(x-2)=0

x=3 or x=-5/3 or x=2

However, as @Mark D points out, all the solutions but x=3 give negative numbers within the square roots and so only x=3 is valid:

sqrt(x^2-x-6)

sqrt(2^2-2-6)=>sqrt(4-2-6)=>sqrt(-4)

sqrt((-5/3)^2+5/3-6)=>sqrt(25/9+15/9-54/9)=>sqrt(-14/9)