Roots of a quadratic equation. Given: \(\displaystyle{a}{\left({f{{\left({x}\right)}}}\right)}^{{2}}+{b}{f{{\left({x}\right)}}}+{c}={0}.\)

Deegan Chase

Deegan Chase

Answered question

2022-04-01

Roots of a quadratic equation.
Given: a(f(x))2+bf(x)+c=0.

Answer & Explanation

disolutoxz61

disolutoxz61

Beginner2022-04-02Added 12 answers

The equation a(f(x))2+b(f(x))+c=0 is "quadratic" only if f(x) is itself a linear function. If u is a solution to au2+bu+c=0 (there might be two values of u) then any solution to f(x)=u is solution to a(f(x))2+b(f(x))+c=0. How many that is depends on the function, f.
Nunnaxf18

Nunnaxf18

Beginner2022-04-03Added 18 answers

When you solve the equation, you will get a value for f(x) first. That would be the root: r=b±b24ac2a. Then you would solve f(x)=r for x.

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