Question

Solve for x:

$\sqrt{x^2+x-6}-x+2=\sqrt{x^2-7x+10},xϵR$

Answer 1

$\sqrt{x^2+x-6}-x+2=\sqrt{x^2-7x+10}$

$\sqrt{(x+3)(x-2)}-(x-2)=\sqrt{(x-5)(x-2)}$

$\sqrt{x-2}[\sqrt{x+3}-\sqrt{x-2}-\sqrt{x-5}]=0$

$Either\sqrt{x-2}=0\to x=2$

$\sqrt{x+3}-\sqrt{x-2}=\sqrt{x-5}$

Squaring on both sides

$(x+3)+(x-2)-2\sqrt{x+3}\sqrt{x-2}=x-5$

$x+6=2\sqrt{x+3}\sqrt{x-2}$

Squaring on both sides

$x^2+12x+36=4(x^2+x-6)$

x = 6, x = -$\frac{10}{3}$

While substituting x = 2, 6 and -$\frac{10}{3}$ in the original equation, we find that x = -$\frac{10}{3}$ does not satisfy the equation. [Because the equation involves radical]

Therefore x = 2, 6