Question

please i need the answer urgently​

Answer 1

Given

• An expression having moduls function

To find

roots of the Equation

Solution

।x²-2।²+।x-2।-6 =0

Let x-2 = a____equation

then mode of a be + a

a²+a -6 =0

=>a²+3a-2a-6=0

=>a(a+3)-2(a+3) =0

=>(a+3) (a-2)=0

=>a=-3,2,0

Now putting the value of a in Equation

x-2=-3

X=-1

and

=>x-2=2

X=4

Here putting the value of -1 in actual Equation they not satisfied

Therefore value of X= (0,4)