Question

2- rootunder5 / 2+ rootunder5 = a rootunder5 + b . Find the value of a and b.​

Answer 1

Question:

Find the value of a and b if,

$\sf \dfrac{2 - \sqrt{5} }{2 + \sqrt{5}} = a \sqrt{5} + b$

Answer:

The values are: a is 4 and b is (-9).

Step-by-step explanation:

$\longrightarrow \: \sf \dfrac{2 - \sqrt{5} }{2 + \sqrt{5}} = a \sqrt{5} + b$

Rationalize the denominator.

$\rightarrow \: \sf \dfrac{(2 - \sqrt{5}) \times (2 - \sqrt{5}) }{2 + (\sqrt{5}) \times (2 - \sqrt{5})} = a \sqrt{5} + b$

$\rightarrow \sf \dfrac{(2 - \sqrt{5})^{2} }{ {2}^{2} - (2)( \sqrt{5}) + (2)( \sqrt{5} ) - { (\sqrt{5}) }^{2}} = a \sqrt{5} + b$

$\rightarrow \: \sf \dfrac{4 - 4 \sqrt{5} + 5 }{4 - 2 \sqrt{5} + 2 \sqrt{5} - 5} = a \sqrt{5} + b$

$\rightarrow \: \sf \dfrac{4 - 4 \sqrt{5} + 5 }{4 - 5} = a \sqrt{5} + b$

$\rightarrow \: \sf \dfrac{4 - 4 \sqrt{5} + 5 }{ - 1} = a \sqrt{5} + b$

$\rightarrow \: \sf { - 4 + 4 \sqrt{5} - 5 } = a \sqrt{5} + b$

$\rightarrow \: \sf { 4\sqrt{5} - 9 } = a \sqrt{5} + b$

$\rightarrow \: \sf { \red{4}\sqrt{5} + \blue{( - 9 )}} = \red{a} \sqrt{5} + \blue{b}$

Here, we can see that,

• a = 4
• b = (-9)

Therefore, a is 4 and b is (-9).

Answer 2

Answer:

$\huge \sf \red{a = 4} \\ \huge \sf \blue{b = - 9}$

Step-by-step explanation:

$\sf \frac{2 - \sqrt{5} }{2 + \sqrt{5} } = a \sqrt{5} + b \\ \\ = > \sf \frac{2 - \sqrt{5} }{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } = a\sqrt{5} + b \\ \\ = > \sf \frac{(2 - \sqrt{5} {)}^{2} }{ {2}^{2} - { \sqrt{5} }^{2} } = a\sqrt{5} + b \\( \sf \: using \: formula(a + b)(a - b) = {a}^{2} - {b}^{2} ) \\ \\ = > \sf \frac{ {2}^{2} - (2)(2)( \sqrt{5}) + \sqrt{ {5}^{2} } }{4 - 5} = a\sqrt{5} + b \\ ( \sf \: using \: formula(a - b {)}^{2} = {a}^{2} - 2ab - {b}^{2} ) \\ \\ = > \sf\frac{4 - 4 \sqrt{5} + 5 }{ - 1} = a\sqrt{5} + b \\ \\ = > \sf \frac{ - 4 \sqrt{5} + 9 }{ - 1} = a \sqrt{5} + b \\ \\ = > \sf4 \sqrt{5} - 9 = a\sqrt{5} + b \\ \sf {\: hence \: {a = 4 \: and \: b = - 9}}$

hope it helps.