Question

Given the quadratic function f(x) = 4x^2 - 4x + 3, determine all possible solutions for f(x) = 0

Answer 1

Answer:

The solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:4x^2\:-\:4x\:+\:3$

Let us determine all possible solutions for f(x) = 0

$0=4x^2-4x+3$

switch both sides

$4x^2-4x+3=0$

subtract 3 from both sides

$4x^2-4x+3-3=0-3$

simplify

$4x^2-4x=-3$

Divide both sides by 4

$\frac{4x^2-4x}{4}=\frac{-3}{4}$

$x^2-x=-\frac{3}{4}$

Add (-1/2)² to both sides

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{3}{4}+\left(-\frac{1}{2}\right)^2$

$x^2-x+\left(-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\left(x-\frac{1}{2}\right)^2=-\frac{1}{2}$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solving

$x-\frac{1}{2}=\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=\sqrt{-1}\sqrt{\frac{1}{2}}$                 ∵ $\sqrt{-\frac{1}{2}}=\sqrt{-1}\sqrt{\frac{1}{2}}$

as

$\sqrt{-1}=i$

so

$x-\frac{1}{2}=i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2}$

also solving

$x-\frac{1}{2}=-\sqrt{-\frac{1}{2}}$

$x-\frac{1}{2}=-i\sqrt{\frac{1}{2}}$

Add 1/2 to both sides

$x-\frac{1}{2}+\frac{1}{2}=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

$x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$

Therefore, the solutions to the quadratic function are:

$x=i\sqrt{\frac{1}{2}}+\frac{1}{2},\:x=-i\sqrt{\frac{1}{2}}+\frac{1}{2}$