Question

If f(x) = 2x2 - 1 and g(x) = x2 + x + 1. for how many integral values of x is f (x - 1 = g (x + 1)2
0 0
0 1
O 2
More than 2​

Answer 1

Given:

$\bold{f(x)= 2x^2-1}\\\\\bold{g(x)=x^2+x+1}$

To find:

$x= ?\\when, f(x-1)=g(x+1)^2$

Solution:

${f(x)= 2x^2-1}\....(i)\\\\{g(x)=x^2+x+1}.....(i)\\\\\ in \ equation (i) \ we \ subtract \ from 1 \\ \\\ Equation: \\\\ f(x)= 2x^2-1 \\\\ \ subtract \ from 1$

$f((x)-1) = 2x^2 -1 -1\\\\f(x-1)= 2x^2-2\\\\\ in \ equation (ii) \ we \ add \ the \ value \ then \ square \ the\ equation:$

$\ equation:\\\\g(x)= x^2+x+1\\\\\ add \ 1\\\\g((x)+1)= x^2+x+1+1\\\\g(x+1)=x^2+x+2\\\\\ square \ the \ equation:\\\\g(x+1)^2= (x^2+x+2)^2 \\\\Formula:\\\\\bold{(a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ca}$

$g(x+1)^2= x^4+x^2+4+2x^3+4x+4x^2$

$\ f(x-1)=g(x+1)^2$

add the calculate value in above equation:

$\Rightarrow x^2+x+2= x^4+x^2+4+2x^3+4x+4x^2\\\\\Rightarrow x^2+x+2= x^4+2x^3+5x^2+4x+4\\\\\Rightarrow x^4+2x^3+4x^2+3x+2$

if we put the value 1, 2, 3, in place of x it will give more then 2 value That's why the answer is more then 2