How many real numbers $x$ are solutions to the equation $|x - 2| - 4 =\frac{1}{|x - 3|}$ ?
Problem
Source:
Tags: algebra, equation
biscuit02
22.09.2021 14:30
For $x<2$
$(x+2)(x-3)=1$
$x^2-x-7=0$
Only valid solution is $x=\frac{1-\sqrt{29}}{2}$
For $2 \leq x \leq 3$
$(x-6)(x-3)=-1$
$(x-5)(x-4)=0$
No valid solutions
For $x\geq 3$
$(x-6)(x-3)=1$
$x^2-9x+17=0$
$x= \frac{9+\sqrt{13}}{2}$
Total valid solutions = $\boxed{2}$
Maitreyi9
22.09.2021 17:09
There can be 3 cases Case 1 When,x<2 Then, We can write (3-x){( 2-x)-4}=1 On equating we will get x^2 -x -7=0 By using shreedhacharya x= (1+under root 1+28)/2= (1+under root 29)/2 Or (1-under root 1+28)/2= (1-under root 29)/2 So we get 2 solutions here Case 2 When,2<=x<3 (3-x)[(x-2)-4]=1 Then we get x^2 - 9x + 19=0 From here we would be getting imaginary roots of x and no real roots Case 3 When,3<=x (x-3)[(x-2)-4]=1 By solving again we get x=(9+under root 13)/2 Or x=(9-under root 13)/2 We get 2 real values of x from case 3 Hence, there are 4 real no. of x as the solution for the equation