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Troyanec [42]
2 years ago
15

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

Mathematics
1 answer:
solong [7]2 years ago
4 0

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}

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