Question

find the square root
8-√15
please explain,no spamming​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that,

$\rm :\longmapsto\: \sqrt{8 - \sqrt{15} } = \sqrt{x} - \sqrt{y} \: (where \: x > y) - - (1)$

On squaring both sides, we get

$\rm :\longmapsto\:8 - \sqrt{15} = {( \sqrt{x} - \sqrt{y} )}^{2}$

$\rm :\longmapsto\:8 - \sqrt{15} = {( \sqrt{x} )}^{2} + {( \sqrt{y} )}^{2} - 2 \sqrt{x} \sqrt{y}$

$\rm :\longmapsto\:8 - \sqrt{15} = x + y - 2 \sqrt{xy}$

can also be rewritten as

$\rm :\longmapsto\:8 - \sqrt{15} = x + y - \sqrt{4xy}$

On comparing, we get

$\rm :\longmapsto\:x + y = 8 - - - (2)$

$\rm :\longmapsto\:4xy = 15 - - - (3)$

We know that,

$\rm :\longmapsto\: {(x - y)}^{2} = {(x + y)}^{2} - 4xy$

$\rm :\longmapsto\: {(x - y)}^{2} = {(8)}^{2} -15$

$\rm :\longmapsto\: {(x - y)}^{2} = 64 -15$

$\rm :\longmapsto\: {(x - y)}^{2} = 49$

$\rm :\longmapsto\: {(x - y)}^{2} = {7}^{2}$

$\rm :\longmapsto\:x - y = 7 - - - (4) \: \: \: \: as \: x > y$

On adding equation (2) and equation (4), we get

$\rm :\longmapsto\:x + y + x - y = 8 + 7$

$\rm :\longmapsto\:2x = 15$

$\bf\implies \:x = \dfrac{15}{2}$

On Subtracting equation (4) from equation (2), we get

$\rm :\longmapsto\:x + y - x + y = 8- 7$

$\rm :\longmapsto\:2y = 1$

$\bf\implies \:y= \dfrac{1}{2}$

So, on substituting the values of x and y in equation (1), we get

$\underbrace{\boxed{ \tt{ \: \bf \: \sqrt{8 \: - \: \sqrt{15} } \: = \: \sqrt{\bigg(\dfrac{15}{2} \bigg)} \: - \: \sqrt{\bigg(\dfrac{1}{2} \bigg)}}}}$

Alternative Method :-

$\rm :\longmapsto\: \sqrt{8 - \sqrt{15} }$

can be rewritten as

$\rm \: = \: \: \: \sqrt{\bigg(\dfrac{16}{2} \bigg) - \sqrt{15} }$

$\rm \: = \: \: \: \sqrt{\bigg(\dfrac{15 + 1}{2} \bigg) - \sqrt{15} }$

$\rm \: = \: \: \: \sqrt{\bigg(\dfrac{15}{2} \bigg) + \bigg(\dfrac{1}{2} \bigg) - 2 \sqrt{\bigg(\dfrac{15}{2 \times 2} \bigg)} }$

$\rm \: = \: \: \: \sqrt{ \bigg( { \sqrt{\bigg(\dfrac{15}{2} \bigg)} } \bigg)^{2} + {\bigg( \sqrt{\bigg(\dfrac{1}{2} \bigg)} \bigg) }^{2} - 2 \sqrt{\bigg(\dfrac{15}{2} \bigg)} \sqrt{\bigg(\dfrac{1}{2} \bigg)} }$

$\rm \: = \: \: {\bigg( \sqrt{\bigg(\dfrac{15}{2} \bigg)} - \sqrt{\bigg(\dfrac{1}{2} \bigg)} } \bigg)^{2}$

$\rm \: = \: \: { \sqrt{\bigg(\dfrac{15}{2} \bigg)} - \sqrt{\bigg(\dfrac{1}{2} \bigg)} }$

Hence,

$\underbrace{\boxed{ \tt{ \: \bf \: \sqrt{8 \: - \: \sqrt{15} } \: = \: \sqrt{\bigg(\dfrac{15}{2} \bigg)} \: - \: \sqrt{\bigg(\dfrac{1}{2} \bigg)}}}}$

Additional Information :-

More Identities to know

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)