Answer: The value of 'x' is -1.8 where the given equation approximately equal
Step-by-step explanation:
It is given that the value of 'x' lies in the range of -2 to -1.5 and we need to find out the value of 'x' where both the equations become equal
Given set of equations:
[tex]y=\frac{1}{(x+2)}[/tex] .....(1)
[tex]y=x^2+2[/tex] .....(2)
Plugging value of 'y' from equation 2 to equation 1:
[tex]x^2+2=\frac{1}{(x+2)}\\\\x^2+2(x+2)=1\\\\x^3+2x^2+2x+4=1\\\\x^3+2x^2+2x+3=0[/tex]
On solving, the real root comes out to be, x = -1.8
Hence, the value of 'x' is -1.8 where the given equation approximately equal
