Question

√x^2-16 - √x^2-8x + 16 = √x^2 -5x +4
Can anyone pls solve it with explanation??

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: \sqrt{ {x}^{2} - 16 } - \sqrt{ {x}^{2} - 8x + 16} = \sqrt{ {x}^{2} - 5x + 4}$

can be rewritten as

$\rm :\longmapsto\: \sqrt{ {x}^{2} - 16 } - \sqrt{ {x}^{2} - 8x + 16} - \sqrt{ {x}^{2} - 5x + 4} = 0$

$\rm :\longmapsto\: \sqrt{ {x}^{2} - {4}^{2}}- \sqrt{{x}^{2}- 4x - 4x + 16} - \sqrt{ {x}^{2} - 4x - x + 4} = 0$

$\rm :\longmapsto\: \sqrt{x + 4)(x - 4)} - \sqrt{x(x - 4) - 4(x - 4)} - \sqrt{x(x - 4) - 1(x - 4)} = 0$

$\rm :\longmapsto\: \sqrt{(x + 4)(x - 4)} - \sqrt{(x - 4)(x - 4)} - \sqrt{(x - 4)(x - 1)} = 0$

$\rm :\longmapsto\: \sqrt{x - 4}( \sqrt{x + 4} - \sqrt{x - 4} - \sqrt{x - 1}) = 0$

$\rm :\implies\:\: \sqrt{x - 4} = 0 \: or \: \sqrt{x + 4} - \sqrt{x - 4} - \sqrt{x - 1} = 0$

$\bf\implies \:x - 4 = 0 \: \: \: or \: x = 4$

Also,

$\rm :\longmapsto\: \sqrt{x + 4} - \sqrt{x - 4} - \sqrt{x - 1} = 0$

can be rewritten as

$\rm :\longmapsto\: \sqrt{x + 4} - \sqrt{x - 4} = \sqrt{x - 1}$

On squaring both sides, we get

$\rm :\longmapsto\: (\sqrt{x + 4} - \sqrt{x - 4})^{2} = (\sqrt{x - 1})^{2}$

$\rm :\longmapsto\:x + 4 + x - 4 - 2\sqrt{x + 4} \sqrt{x - 4} = x - 1$

$\rm :\longmapsto\:2x - 2\sqrt{ {x}^{2} - 16} = x - 1$

$\rm :\longmapsto\: - 2\sqrt{ {x}^{2} - 16} = x - 1 - 2x$

$\rm :\longmapsto\: - 2\sqrt{ {x}^{2} - 16} = - x - 1$

$\rm :\longmapsto\: - 2\sqrt{ {x}^{2} - 16} = - (x + 1)$

$\rm :\longmapsto\: 2\sqrt{ {x}^{2} - 16} = (x + 1)$

On squaring both sides, we get

$\rm :\longmapsto\:4( {x}^{2} - 16) = {(x + 1)}^{2}$

$\rm :\longmapsto\:4{x}^{2} - 64 = {x}^{2} + 1 + 2x$

$\rm :\longmapsto\:4{x}^{2} - 64 - {x}^{2} - 1 - 2x = 0$

$\rm :\longmapsto\:3{x}^{2} - 2x- 65 = 0$

$\rm :\longmapsto\:3{x}^{2} - 15x + 13x- 65 = 0$

$\rm :\longmapsto\:3x(x - 5) + 13(x - 5) = 0$

$\rm :\longmapsto\:(x - 5)(3x + 13) = 0$

$\bf\implies \:x = 5 \: \: or \: \: x = - \: \dfrac{13}{3}$

So, Solution is

$\bf\implies \:x = 5 \: \: or \: \: x = - \: \dfrac{13}{3} \: \: or \: \: x = 4$