Question

sqrt 7-1 sqrt 7 +1 - sqrt 7 +2 sqrt 7 -1 =a+b sqrt 7​

Answer 1

Step-by-step explanation:

Correct Question:—

$\frac{ \sqrt{7} - 1}{ \sqrt{7} + 1 } - \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1 } = a + b \sqrt{7}$

To find out:—

• The value of a and b in expression.

Solution:—

Let's solve the problem

we have,

$\frac{ \sqrt{7} - 1}{ \sqrt{7} + 1 } - \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1 }$

$⟹ \frac{ \sqrt{7} - 1 }{{7} + 1 } \times \frac{ \sqrt{7} - 1 }{ \sqrt{7} - 1} - \frac{ \sqrt{7} + 1}{ \sqrt{7} - 1} \times \frac{ \sqrt{7} + 1}{ \sqrt{7} + 1}$

$⟹ \frac{( \sqrt{7} - 1)( \sqrt{7} - 1)}{( \sqrt{ 7 } + 1)( \sqrt{7} - 1)} - \frac{( \sqrt{7} + 1)( \sqrt{7} + 1) }{( \sqrt{7} - 1)( \sqrt{7} + 1) }$

$⟹ \frac{( \sqrt{7} - 1 {)}^{2} }{( \sqrt{7} {)}^{2} - (1 {)}^{2} } - \frac{( \sqrt{7} + 1 {)}^{2} }{( \sqrt{7} {)}^{2} - (1 {)}^{2} }$

$⟹ \frac{( \sqrt{7} {)}^{2} + (1 {)}^{2} - 2( \sqrt{7} )(1)}{7 - 1} - \frac{( \sqrt{7} {)}^{2} + (1 {)}^{2} + 2( \sqrt{7})(1) }{7 - 1}$

$⟹ \frac{7 + 1 - 2 \sqrt{7} }{6} - \frac{7 + 1 + 2 \sqrt{7} }{6}$

$⟹ \frac{8 - 2 \sqrt{7} }{6} - \frac{8 + 2 \sqrt{7} }{6}$

$⟹ \frac{1}{6} (8 - 2 \sqrt{7} - 8 - 2 \sqrt{7} )$

$⟹0 - \frac{4 \sqrt{7} }{6}$

$⟹0 + \bigg( \frac{ - 2}{3} \bigg) \sqrt{7}$

$∴\: \: 0 + \bigg( \frac{ - 2}{3} \bigg) \sqrt{7} = a + b \sqrt{7}$

On comparing the value of ;

$a = 0 \: \: and \: \: b = - \frac{2}{3}$

Answer:—

$\mathrm{Hence, the \: value \: of \: a = 0 \: and \: b = - \frac{2}{3} .}$

Used formulae:—

• (a-b)(a-b) = (a-b)² = a² + b² - 2ab
• (a+b)(a+b) = (a+b)² = a²+b²+2ab
• (a+b)(a-b) = a² - b²
• (a-b)(a+b) = a² - b²

Learn more;

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