If f(x)=2x−1 and 6f(x2)−f2(x)=6x2−x are satisfied by x1 and x2, then the value of x1+x2= ....
Explanation
Given the function f(x)=2x−1. We want to find the value of x1+x2 that satisfies the equation 6f(x2)−f2(x)=6x2−x.
First, we find the forms of f(x2) and f2(x) or (f(x))2.
f(x2)=2(x2)−1=2x2−1
f2(x)=(2x−1)2=4x2−4x+1
Next, we substitute these results into the given equation.
6f(x2)−f2(x)=6x2−x
6(2x2−1)−(4x2−4x+1)=6x2−x
12x2−6−4x2+4x−1=6x2−x
8x2+4x−7=6x2−x
We move all terms to the left side to form a quadratic equation ax2+bx+c=0.
8x2−6x2+4x+x−7=0
2x2+5x−7=0
From the quadratic equation 2x2+5x−7=0, we get a=2, b=5, and c=−7.
The sum of the roots of the quadratic equation (x1+x2) can be calculated using the formula −ab.
x1+x2=−25
So, the value of x1+x2 is −25.