Question

$\huge \bf Question$
Brain Treasure
Solve it :-
$\tt \: 2(x + \frac{1}{x}) {}^{2} = 10 - x - \frac{1}{x}$
Also verify it
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Answer 1

Answer,

x = 1 , 1 , -2 , -0.5

Given,

$\sf 2(x+\dfrac{1}{x})^2=10-x-\dfrac{1}{x}$

Solution,

$\sf 2(x+\dfrac{1}{x})^2=10-x-\dfrac{1}{x} \\\\ 2(x^2+(\dfrac{1}{x})^2+2(x)(\dfrac{1}{x}))=10-x-\dfrac{1}{x} \\\\ 2(x^2+\dfrac{1}{x^2}+2)=\dfrac{10x-x^2-1}{x} \\\\ 2x^2+\dfrac{2}{x^2}+4=\dfrac{10x-x^2-1}{x} \\\\ x(2x^2+\dfrac{2}{x^2}+4)=10x-x^2-1 \\\\ 2x^3+\dfrac{2x}{x^2}+4x=10x-x^2-1 \\\\ 2x^3+\dfrac{2}{x}+4x=10x-x^2-1 \\\\ \dfrac{2x^4+2+4x^2}{x}=10x-x^2-1 \\\\ 2x^4+2+4x^2=x(10x-x^2-1) \\\\ 2x^4+2+4x^2=10x^2-x^3-x \\\\ 2x^4+x^3+4x^2-10x^2+x+2=0 \\\\ 2x^4+x^3-6x^2+x+2=0$

Thus, we got a polynomial of degree 4.

we have to find the factors of the polynomial.

By trial and error method,

Put x = 1,

2(1)⁴ + 1³ - 6(1)² + 1 + 2 = 0

2(1) + 1 - 6 + 3 = 0

2 + 1 - 6 + 3 = 0

3 - 6 + 3 = 0

 0 = 0

Therefore, 1 is a zero of the polynomial.

(x - 1) is a factor

Now, divide the polynomial by (x - 1)

$\Large \begin{array}{c|c|c}\sf x-1&\sf \quad 2x^4+x^3-6x^2+x+2&\sf 2x^3+3x^2-3x-2\\&\sf 2x^4-2x^3 \qquad \qquad \quad \\ \cline{2-2} & \sf \qquad \ \ 3x^3-6x^2+x+2 & \\ & \sf 3x^3-3x^2 \ \\ \cline{2-2} & \sf \qquad \qquad \quad -3x^2+x+2 & \\ & \sf \qquad \qquad \quad -3x^2+3x \ \ \\ \cline{2-2} & \sf \qquad \qquad \quad \: -2x+2 & \\ & \sf \qquad \qquad \quad -2x+2 \\ \cline{2-2} & \sf \qquad \qquad \quad 0 \end{array}$

Quotient = 2x³ + 3x² - 3x - 2

Again, by trial and error method,

Put x = 1,

= 2(1)³ + 3(1)² - 3(1) - 2

= 2(1) + 3(1) - 3 - 2

= 2 + 3 - 3 - 2

= 0

1 is a zero of the polynomial.

Therefore, (x - 1) is also a factor of 2x³ + 3x² - 3x - 2

Divide 2x³ + 3x² - 3x - 2 by (x-1)

$\Large \begin{array}{c|c|c}\sf x-1&\sf 2x^3+3x^2-3x-2&\sf 2x^2+5x+2\\&\sf \ \ 2x^3-2x^2 \qquad \qquad \quad \\ \cline{2-2} & \sf \qquad \ \ 5x^2-3x-2 & \\ & \sf \quad 5x^2-5x \ \\ \cline{2-2} & \sf \qquad \qquad \quad 2x-2 & \\ & \sf \qquad \qquad \quad 2x-2 \\ \cline{2-2} & \sf \qquad \qquad \quad 0 \end{array}$

Quotient = 2x² + 5x + 2

Now, factorize 2x² + 5x + 2

2x² + 5x + 2

2x² + 4x + x + 2

2x(x + 2) + 1(x + 2)

(x + 2) (2x + 1)

Therefore, (x + 2) and (2x + 1) are factors

• x + 2 = 0 ; x = -2
• 2x + 1 = 0 ; x = -1/2 = -0.5

x = 1 , 1 , -2 , -0.5

Verification :

Put x = 1,

$\sf 2(1+\dfrac{1}{1})^2=10-1-\dfrac{1}{1} \\\\ 2(1+1)^2=10-1-1 \\\\ 2(2)^2=10-2 \\\\ 2(4)=8\\\\8=8$

Put x = -2,

$\sf 2(-2+\dfrac{1}{-2})^2=10-(-2)-\dfrac{1}{-2} \\\\ 2(-2-0.5)^2=10+2+0.5 \\\\ 2(-2.5)^2=12.5 \\\\ 2(6.25)=12.5 \\\\ 12.5=12.5$

Put x = -0.5,

$\sf 2(-0.5+\dfrac{1}{-0.5})^2=10-(-0.5)-\dfrac{1}{-0.5} \\\\ 2(-0.5-2)^2=10+0.5+2 \\\\ 2(-2.5)^2=12.5 \\\\ 2(6.25)=12.5 \\\\ 12.5=12.5$

Hence verified!