Question

If x2²+1/x² =7,find the values of
3x²–3x²​

Please help me​

Answer 1

Corrected question:

If x² + (1/x²) = 7, find the value/s of the expression 3x² - (3/x²).

Explanation:

ATQ:

$\Longrightarrow \ \sf x^{2} + \dfrac{1}{x^2} = 7 \ \dashrightarrow \ Eq(1)$

We know that;

⇒ (a - b)² = a² + b² - 2ab

⇒ (a - b)² + 2ab = a² + b²

Let "a" = x and "b" = 1/x, Using the identity (a - b)² + 2ab = a² + b² in Equation 1 we get:

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 + \ 2\Bigg\{x\Bigg\}\Bigg\{\dfrac{1}{x}\Bigg\} = 7$

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 + \ 2 = 7$

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 = 7 - 2$

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 = 5$

$\Longrightarrow \ \sf x - \dfrac{1}{x} = \pm \sqrt{5} \ \dashrightarrow Eq(2)$

Similarly, using the identity (a + b)² - 2ab = a² + b² in Equation 1 we get:

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 - \ 2\Bigg\{x\Bigg\}\Bigg\{\dfrac{1}{x}\Bigg\} = 7$

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 - \ 2 = 7$

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 = 7 + 2$

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 = 9$

$\Longrightarrow \ \sf x + \dfrac{1}{x} = \sqrt{9}$

$\Longrightarrow \ \sf x + \dfrac{1}{x} = \pm 3 \ \dashrightarrow Eq(3)$

According to the question, we need to find the value of 3x² - (3/x²).

$\Longrightarrow \ \sf 3x^{2} - \dfrac{3}{x^2}$

Take 3 out as common:

$\Longrightarrow \ \sf 3\Bigg\{x^{2} - \dfrac{1}{x^2}\Bigg\}$

Let "a" = x and "b" = 1/x. We know that:

⇒ a² - b² = (a + b)(a - b)

Using this identity we get:

$\Longrightarrow \ \sf 3\Bigg\{x - \dfrac{1}{x}\Bigg\}\Bigg\{x + \dfrac{1}{x}\Bigg\}$

Substitute the values in Eq(2) and Eq(3) in the above equation.

$\Longrightarrow \ \sf 3\Bigg\{\pm \sqrt{5}\Bigg\}\Bigg\{\pm 3\Bigg\}$

$\Longrightarrow \ \sf \pm 9\sqrt{5}$

Therefore the answer is ± 9√5.

Answer 2

Answer:

Corrected question:

If x² + (1/x²) = 7, find the value/s of the expression 3x² - (3/x²).

Explanation:

ATQ:

$\Longrightarrow \ \sf x^{2} + \dfrac{1}{x^2} = 7 \ \dashrightarrow \ Eq(1)$

We know that;

⇒ (a - b)² = a² + b² - 2ab

⇒ (a - b)² + 2ab = a² + b²

Let "a" = x and "b" = 1/x, Using the identity (a - b)² + 2ab = a² + b² in Equation 1 we get:

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 + \ 2\Bigg\{x\Bigg\}\Bigg\{\dfrac{1}{x}\Bigg\} = 7$

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 + \ 2 = 7$

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 = 7 - 2$

$\Longrightarrow \ \sf \Bigg\{x - \dfrac{1}{x}\Bigg\}^2 = 5$

$\Longrightarrow \ \sf x - \dfrac{1}{x} = \pm \sqrt{5} \ \dashrightarrow Eq(2)$

Similarly, using the identity (a + b)² - 2ab = a² + b² in Equation 1 we get:

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 - \ 2\Bigg\{x\Bigg\}\Bigg\{\dfrac{1}{x}\Bigg\} = 7$

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 - \ 2 = 7$

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 = 7 + 2$

$\Longrightarrow \ \sf \Bigg\{x + \dfrac{1}{x}\Bigg\}^2 = 9$

$\Longrightarrow \ \sf x + \dfrac{1}{x} = \sqrt{9}$

$\Longrightarrow \ \sf x + \dfrac{1}{x} = \pm 3 \ \dashrightarrow Eq(3)$

According to the question, we need to find the value of 3x² - (3/x²).

$\Longrightarrow \ \sf 3x^{2} - \dfrac{3}{x^2}$

Take 3 out as common:

$\Longrightarrow \ \sf 3\Bigg\{x^{2} - \dfrac{1}{x^2}\Bigg\}$

Let "a" = x and "b" = 1/x. We know that:

⇒ a² - b² = (a + b)(a - b)

Using this identity we get:

$\Longrightarrow \ \sf 3\Bigg\{x - \dfrac{1}{x}\Bigg\}\Bigg\{x + \dfrac{1}{x}\Bigg\}$

Substitute the values in Eq(2) and Eq(3) in the above equation.

$\Longrightarrow \ \sf 3\Bigg\{\pm \sqrt{5}\Bigg\}\Bigg\{\pm 3\Bigg\}$

$\Longrightarrow \ \sf \pm 9\sqrt{5}$

Therefore the answer is ± 9√5.