Question

if √7+4/√7-4=m+n√7 find the value of 'm' and 'n'​

Answer 1

Answer:

√7+4/√7-4=m+n√7

by rationalising the denominator

(√7+4)²/7-16= m+n√7

7+16+8√7/-9=m+n√7

23+8√7/-9=m+n√7

-23/9-8√7/9=m+n√7

m=-23/9,n=-8/9

Answer 2

Answer:

• m = - 23/9, n = - 8/9

Step-by-step explanation:

As per information provided in the question, We have:

√7 + 4/√7 - 4 = m + n√7

We are asked to find the values of m and n.

In order to find the values of m and n, We need to Rationalize √7 + 4/√7 - 4 until it will be in this form ⇒ m + n√7.

Rationalising √7 + 4/√7 - 4,

$\longmapsto \rm \dfrac{ \sqrt{7} + 4 }{ \sqrt{7} - 4}$

Multiplying the fraction with the rationalising factor of the denominator,

$\longmapsto \rm \dfrac{ \sqrt{7} + 4 }{ \sqrt{7} - 4} \times \dfrac{ \sqrt{7} + 4 }{ \sqrt{7} + 4 }$

Rearranging the terms.

$\longmapsto \rm \dfrac{( \sqrt{7} + 4 )( \sqrt{7} + 4) }{ (\sqrt{7} - 4)( \sqrt{7} + 4) }$

$\longmapsto \rm \dfrac{ {( \sqrt{7} + 4 )}^{2} }{ (\sqrt{7} - 4)( \sqrt{7} + 4) }$

Using the identity - (a - b)(a + b) = a² - b² in denominator & Using - (a + b)² = (a)² + (b)² + 2ab in numerator,

$\longmapsto \rm \dfrac{ {( \sqrt{7} + 4 )}^{2} }{ ( {\sqrt{7})}^{2} - {(4)}^{2} }$

$\longmapsto \rm \dfrac{{( \sqrt{7} )}^{2} + {(4)}^{2} + 2\sqrt{7} \times 4}{ ( {\sqrt{7})}^{2} - {(4)}^{2} }$

$\longmapsto \rm \dfrac{ 28 + 8 \sqrt{7} }{ 7 -16 }$

$\longmapsto \rm \dfrac{ 23 + 8 \sqrt{7} }{ - 9 }$

$\longmapsto \rm - \dfrac{ 23 + 8 \sqrt{7} }{ 9 }$

Thus, The value of m is - 23/9 & the value of n is - 8/9.