Question

A diferença entre a maior e a menor raiz real da equação (2x –+ 1) ∙ (2x –– 1) = 4x –é igual a:​

Answer 1

Explicação passo-a-passo:

2x−

2

+1)∗(2x−

2

−1)=4x−

8

4x

2

−2x

2

−2x−2x

2

+(

2

)

2

+

2

+2x−

2

−1=4x−

8

4x

2

−2x

2

−̸2x−2x

2

+2+̸

̸2

+̸2x−̸

̸2

−1=4x−

8

4x

2

−2x

2

−2x

2

+2−1=4x−

8

4x

2

−4x

2

+1=4x−

8

\begin{lgathered}4x^2 - 4x \sqrt{2} + 1 -4x + 2\sqrt{2} \\ \\ \\ 4x^2 - x (4 + 4 \sqrt{2}) +1 + 2 \sqrt{2} \\ \\ \\\end{lgathered}

4x

2

−4x

2

+1−4x+2

2

4x

2

−x(4+4

2

)+1+2

2

=====

Calculando as raízes igualamos os termos à zero

\begin{lgathered}2x - 1 = 0 \\ \\ x' = \dfrac{1}{2} \\ \\ \\ 2x -(1 + 2 \sqrt{2}) \\ \\ 2x = (1 + 2 \sqrt{2}) \\ \\ \\ x'' = \dfrac{(1 + 2 \sqrt{2})}{2}\end{lgathered}

2x−1=0

x

′

=

2

1

2x−(1+2

2

)

2x=(1+2

2

)

x

′′

=

2

(1+2

2

)

X' - X''

\begin{lgathered}\dfrac{1}{2} - \dfrac{(1 + 2 \sqrt{2})}{2} \\ \\ \\ => - \sqrt{2}\end{lgathered}

2

1

−

2

(1+2

2

)

=>−

2

Answer 2

(2x-1)^2 = 4x

4x^2-4x+1=4x

4x^2-4x-4x=-1

4x^2-8x=-1

x^2-2x=-1/4

x^2-2x+1^2=-1/4+1^2

(x-1)^2 = 3/4

x' = $-\frac{\sqrt[]{3}}{2} + 1$

x'' = $\frac{\sqrt{3} }{2}  + 1$

x' - x'' = $\sqrt{3}$