Question

Solve the modulus function:
|x²-2| = 2|x-3|

Answer 1

Step-by-step explanation:

We have,

$| {x}^{2} - 2 | = 2 |x - 3|$

$\implies | {x}^{2} - 2 |^{2} = (2 |x - 3| )^{2}$

$\implies ( {x}^{2} )^{2}+ (2)^{2} - 4 {x}^{2} = 4 \{ (x)^{2} + ( 3 )^{2} - 6x \}$

$\implies {x}^{4}+ 4 - 4 {x}^{2} = 4 x^{2} + 36 - 24x$

$\implies {x}^{4} - 32 - 8 {x}^{2} + 24x = 0$

$\implies {x}^{4} - 8 {x}^{2} + 24x - 32 = 0$

$\implies {x}^{4} - 2{x}^{3} + 2{x}^{3} -4{x}^{2} - 4 {x}^{2} + 8x + 16x - 32 = 0$

$\implies {x}^{3}(x - 2) + 2{x}^{2}(x -2) - 4x (x - 2) + 16(x - 2) = 0$

$\implies (x - 2)( {x}^{3} + 2{x}^{2} - 4x+ 16) = 0$

$\implies (x - 2)( {x}^{3} + 4{x}^{2} - 2 {x}^{2} - 8x + 4x+ 16) = 0$

$\implies (x - 2) \{ {x}^{2} (x + 4) - 2x(x + 4)+ 4(x+ 4)\} = 0$

$\implies (x - 2) (x + 4)( {x}^{2} - 2x+ 4) = 0$

$\implies (x - 2) (x + 4)( {x}^{2} - 2x+ 4) = 0$

Here, $x^2 -2x+4$ is always positive becaise its $a>0\:\: \& \:\: D<0$

so, the required values of x are $x=2\:\: \&\:\: x=-4$