Question

please do this I am give you 100 points ​

Answer 1

QUESTION:

x² + 1 = 4x. Find x² + 1/x² and x² - 1/x²

ANSWER:

$\sf x^2 + \dfrac{1}{x^2}= \dfrac{2( {8x}^{2} -4x +1)}{4x - 1}$

$\sf x^2 - \dfrac{1}{x^2}= \dfrac{ 8x(2x - 1)}{4x - 1}$

GIVEN:

• x² + 1 = 4x.

TO FIND:

• x² + 1/x² and x² - 1/x²

EXPLANATION:

x² + 1 = 4x

x² = 4x - 1

$\sf x^2 + \dfrac{1}{x^2}=4x - 1 + \dfrac{1}{4x - 1}$

$\sf x^2 + \dfrac{1}{x^2}= \dfrac{(4x - 1 {)}^{2} +1}{4x - 1}$

$\boxed{\bold{\large{\gray{(A - B)^2 = A^2 - 2AB+B^2}}}}$

$\sf x^2 + \dfrac{1}{x^2}= \dfrac{ {16x}^{2} - 2(4x)(1) +1 +1}{4x - 1}$

$\sf x^2 + \dfrac{1}{x^2}= \dfrac{ {16x}^{2} -8x +2}{4x - 1}$

$\sf x^2 + \dfrac{1}{x^2}= \dfrac{2( {8x}^{2} -4x +1)}{4x - 1}$

$\sf x^2 - \dfrac{1}{x^2}=4x - 1 - \dfrac{1}{4x - 1}$

$\sf x^2 - \dfrac{1}{x^2}= \dfrac{(4x - 1 {)}^{2} - 1}{4x - 1}$

$\boxed{\bold{\large{\gray{(A - B)^2 = A^2 - 2AB+B^2}}}}$

$\sf x^2 - \dfrac{1}{x^2}= \dfrac{ {16x}^{2} - 2(4x)(1) +1 - 1}{4x - 1}$

$\sf x^2 - \dfrac{1}{x^2}= \dfrac{ {16x}^{2} -8x}{4x - 1}$

$\sf x^2 - \dfrac{1}{x^2}= \dfrac{ 8x(2x - 1)}{4x - 1}$

$\sf The\ value\ of \ \left(x^2 + \dfrac{1}{x^2}\right)= \dfrac{2( {8x}^{2} -4x +1)}{4x - 1}$

$\sf The\ value\ of \ \left(x^2 - \dfrac{1}{x^2}\right)= \dfrac{ 8x(2x - 1)}{4x - 1}$

Answer 2

hence the value of (x²+1/x²)=2(8x²-4x+1)/4x-1