Question

If (x² + 1/x²) - 2(x + 1/x) - 6 = 0 and x iѕ rаtіоnаl, thеn x is :​

Answer 1

Solution!!

$\sf \to \left(x^{2}+\dfrac{1}{x^{2}}\right)-2\left(x+\dfrac{1}{x}\right)-6=0$

Remove the unnecessary parentheses.

$\sf \to x^{2}+\dfrac{1}{x^{2}}-2\left(x+\dfrac{1}{x}\right)-6=0$

Distribute -2 through the parentheses.

$\sf \to x^{2}+\dfrac{1}{x^{2}}-2x-\dfrac{2}{x}-6=0$

Write all numerators above the least common denominator x².

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

When the quotient of expressions equals 0, the numerator has to be 0.

$\sf \to x^{4}+1-2x^{3}-2x-6x^{2}=0$

Use the commutative property to reorder the terms.

$\sf \to x^{4}-2x^{3}-6x^{2}-2x+1=0$

Write -2x³ as a difference.

$\sf \to x^{4}+x^{3}-3x^{3}-6x^{2}-2x+1=0$

Write -6x² as a difference.

$\sf \to x^{4}+x^{3}-3x^{3}-3x^{2}-3x^{2}-2x+1=0$

Write -2x as a sum.

$\sf \to x^{4}+x^{3}-3x^{3}-3x^{2}-3x^{2}-3x+x+1=0$

Factor out x³, -3x², -3x and 1 from the expressions.

$\sf \to x^{3}(x+1)-3x^{2}(x+1)-3x(x+1)+1(x+1)=0$

Factor out x + 1 from the expression.

$\sf \to (x+1)(x^{3}-3x^{2}-3x+1)=0$

Use a³ + b³ = (a + b)(a² - ab + b²) to factor the expression.

$\sf \to (x+1)((x+1)(x^{2}-x+1)-3x^{2}-3x)=0$

Factor out -3x from the expression.

$\sf \to (x+1)((x+1)(x^{2}-x+1)-3x(x+1))=0$

Factor out x + 1 from the expression.

$\sf \to (x+1)(x+1)(x^{2}-x+1-3x)=0$

Collect the like terms and write the repeated multiplication in exponential form.

$\sf \to (x+1)^{2}(x^{2}-4x+1)=0$

When the product of factors equals 0, at least one factor is 0.

$\sf \to (x+1)^{2}=0$

$\sf \to (x^{2}-4x+1)=0$

Let's solve.

$\sf \to x+1=0$

$\sf \to x^{2}-4x+1=0$

$\sf$

$\sf \to x=-1$

$\sf \to x=\dfrac{-(-4)\pm \sqrt{(-4)^{2}-4\times 1\times 1}}{2(1)}$

$\sf$

$\sf \to x=-1$

$\sf \to x=\dfrac{4\pm \sqrt{12}}{2}$

$\sf$

$\sf \to x=-1$

$\sf \to x=\dfrac{4\pm 2\sqrt{3}}{2}$

$\sf$

$\sf \to x=-1$

$\sf \to x=\dfrac{4+2\sqrt{3}}{2}$

$\sf \to x=\dfrac{4-2\sqrt{3}}{2}$

$\sf$

$\sf \to x=-1$

$\sf \to x=2+\sqrt{3}$

$\sf \to x=2-\sqrt{3}$

As we can see that we have 3 values of x but it is clearly mentioned in the question that x is rational. Hence, eliminating the second and third values of x as they are irrational, x equals -1 which is rational.

x = -1