Question

If x = square root of (m+n)+square root of (m-n)/square root of (m+n)-square root (m-n), where n which isn't equal 0 , then find the value of nx^2 - 2mx + n.

Answer 1

Answer:

nx² - 2mx + n = 0

Step-by-step explanation:

If x = square root of (m+n)+square root of (m-n)/square root of (m+n)-square root (m-n), where n which isn't equal 0 , then find the value of nx^2 - 2mx + n.

$x = \frac{\sqrt{m + n} + \sqrt{m-n}}{\sqrt{m + n} - \sqrt{m-n}}\\ \\Multiplying \: \& \: divide\: by \: \sqrt{m + n} + \sqrt{m-n}\\ \\x = \frac{\sqrt{m + n} + \sqrt{m-n}}{\sqrt{m + n} - \sqrt{m-n}} \times \frac{\sqrt{m + n} + \sqrt{m-n}}{\sqrt{m + n} + \sqrt{m-n}} \\ \\\\x = \frac{m + n + m - n + 2\sqrt{m^2 - n^2} }{m + n - (m-n) } \\ \\x = \frac{2(m + \sqrt{m^2 - n^2})}{2n} \\ \\x = \frac{m + \sqrt{m^2 - n^2}}{n} \\ \\nx = m + \sqrt{m^2 - n^2} \\ \\nx - m = \sqrt{m^2 - n^2}$

squaring both sides

(nx-m)² = m² - n²

=> n²x² + m² - 2mnx = m² - n²

=> n²x² - 2mnx = - n²

Diving by n both sides

=> nx² - 2mx = - n

=> nx² - 2mx + n = 0

Answer 2

Answer:

$nx^2-2mx+n=0$

Step-by-step explanation:

Given:

$x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}$

$x=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}.\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}+\sqrt{m-n}}$

$x=\frac{(\sqrt{m+n}+\sqrt{m-n})^2}{(\sqrt{m+n})^2-(\sqrt{m-n})^2}$

$x=\frac{(m+n)+(m-n)+2\sqrt{(m+n)(m-n)}}{(\sqrt{m+n})^2-(\sqrt{m-n})^2}$

$x=\frac{2m+2\sqrt{m^2-n^2}}{m+n-m+n}$

$x=\frac{2m+2\sqrt{m^2-n^2}}{2n}$

$x=\frac{m+\sqrt{m^2-n^2}}{n}$

$x^2=\frac{(m+\sqrt{m^2-n^2})^2}{n^2}$

$x^2=\frac{m^2+m^2-n^2+2m\sqrt{m^2-n^2}}{n^2}$

$x^2=\frac{2m^2-n^2+2m\sqrt{m^2-n^2}}{n^2}$

$nx^2=\frac{2m^2-n^2+2m\sqrt{m^2-n^2}}{n^2}.n$

$nx^2=\frac{2m^2-n^2+2m\sqrt{m^2-n^2}}{n}$

$-2mx=-2m(\frac{m+\sqrt{m^2-n^2}}{n})$

Now,

$nx^2-2mx+n$

$=\frac{2m^2-n^2+2m\sqrt{m^2-n^2}}{n}-2m(\frac{m+\sqrt{m^2-n^2}}{n})+n$

$=\frac{2m^2-n^2+2m\sqrt{m^2-n^2}-2m^2-2m\sqrt{m^2-n^2}+n^2}{n}$

$=\frac{0}{n}$

$=0$