find the square root
8-√15
please explain,no spamming
Share
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Let assume that,
[tex]\rm :\longmapsto\: \sqrt{8 - \sqrt{15} } = \sqrt{x} - \sqrt{y} \: (where \: x > y) - - (1)[/tex]
On squaring both sides, we get
[tex]\rm :\longmapsto\:8 - \sqrt{15} = {( \sqrt{x} - \sqrt{y} )}^{2} [/tex]
[tex]\rm :\longmapsto\:8 - \sqrt{15} = {( \sqrt{x} )}^{2} + {( \sqrt{y} )}^{2} - 2 \sqrt{x} \sqrt{y} [/tex]
[tex]\rm :\longmapsto\:8 - \sqrt{15} = x + y - 2 \sqrt{xy} [/tex]
can also be rewritten as
[tex]\rm :\longmapsto\:8 - \sqrt{15} = x + y - \sqrt{4xy} [/tex]
On comparing, we get
[tex]\rm :\longmapsto\:x + y = 8 - - - (2)[/tex]
[tex]\rm :\longmapsto\:4xy = 15 - - - (3)[/tex]
We know that,
[tex]\rm :\longmapsto\: {(x - y)}^{2} = {(x + y)}^{2} - 4xy[/tex]
[tex]\rm :\longmapsto\: {(x - y)}^{2} = {(8)}^{2} -15[/tex]
[tex]\rm :\longmapsto\: {(x - y)}^{2} = 64 -15[/tex]
[tex]\rm :\longmapsto\: {(x - y)}^{2} = 49[/tex]
[tex]\rm :\longmapsto\: {(x - y)}^{2} = {7}^{2} [/tex]
[tex]\rm :\longmapsto\:x - y = 7 - - - (4) \: \: \: \: as \: x > y[/tex]
On adding equation (2) and equation (4), we get
[tex]\rm :\longmapsto\:x + y + x - y = 8 + 7[/tex]
[tex]\rm :\longmapsto\:2x = 15[/tex]
[tex]\bf\implies \:x = \dfrac{15}{2} [/tex]
On Subtracting equation (4) from equation (2), we get
[tex]\rm :\longmapsto\:x + y - x + y = 8- 7[/tex]
[tex]\rm :\longmapsto\:2y = 1[/tex]
[tex]\bf\implies \:y= \dfrac{1}{2} [/tex]
So, on substituting the values of x and y in equation (1), we get
[tex]\underbrace{\boxed{ \tt{ \: \bf \: \sqrt{8 \: - \: \sqrt{15} } \: = \: \sqrt{\bigg(\dfrac{15}{2} \bigg)} \: - \: \sqrt{\bigg(\dfrac{1}{2} \bigg)}}}}[/tex]
Alternative Method :-
[tex]\rm :\longmapsto\: \sqrt{8 - \sqrt{15} } [/tex]
can be rewritten as
[tex]\rm \: = \: \: \: \sqrt{\bigg(\dfrac{16}{2} \bigg) - \sqrt{15} } [/tex]
[tex]\rm \: = \: \: \: \sqrt{\bigg(\dfrac{15 + 1}{2} \bigg) - \sqrt{15} } [/tex]
[tex]\rm \: = \: \: \: \sqrt{\bigg(\dfrac{15}{2} \bigg) + \bigg(\dfrac{1}{2} \bigg) - 2 \sqrt{\bigg(\dfrac{15}{2 \times 2} \bigg)} } [/tex]
[tex]\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)} } [/tex]
[tex]\rm \: = \: \: {\bigg( \sqrt{\bigg(\dfrac{15}{2} \bigg)} - \sqrt{\bigg(\dfrac{1}{2} \bigg)} } \bigg)^{2} [/tex]
[tex]\rm \: = \: \: { \sqrt{\bigg(\dfrac{15}{2} \bigg)} - \sqrt{\bigg(\dfrac{1}{2} \bigg)} }[/tex]
Hence,
[tex]\underbrace{\boxed{ \tt{ \: \bf \: \sqrt{8 \: - \: \sqrt{15} } \: = \: \sqrt{\bigg(\dfrac{15}{2} \bigg)} \: - \: \sqrt{\bigg(\dfrac{1}{2} \bigg)}}}}[/tex]
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)