Question

if 1/root5 - 2=a root 5+b the value of a and b is

Answer 1

Answer:

Value of a is 1 and value of b is 2.

Step-by-step explanation:

According to the given question, we have :

$\longrightarrow \sf { \dfrac{1}{\sqrt{5} - 2} = a \sqrt{5} + b }$

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Rationalising the denominator in L.H.S to find the value of a and b. In order to rationalise the denominator of any fraction, we need to multiply the rationalising factor of the denominator with both the numerator and the denominator of the given fraction.

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In L.H.S :

$\longrightarrow \sf { \dfrac{1}{\sqrt{5} - 2} }$

Here,

• Denominator = $\sf {\sqrt{5}-2 }$

As we know that, rationalising factor of (a - b) is (a + b), so rationalising factor of ($\sf {\sqrt{5}-2 }$) is ($\sf {\sqrt{5} +2 }$). Multiplying ($\sf {\sqrt{5} + 2 }$) with both numerator and denominator.

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$\longrightarrow \sf { \dfrac{1}{\sqrt{5} - 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2} }$

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$\longrightarrow \sf { \dfrac{1(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} }$

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By using identity,

• $\bf {(a +b)(a -b) = a^2 - b^2 }$

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$\longrightarrow \sf { \dfrac{\sqrt{5} + 2}{( \sqrt{5})^2 - (2)^2} }$

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$\longrightarrow \sf { \dfrac{\sqrt{5} + 2}{5 -4} }$

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$\longrightarrow \sf { \dfrac{\sqrt{5} + 2}{1} }$

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$\longrightarrow \sf { \sqrt{5} + 2 }$

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In L.H.S, we get $\sf { \sqrt{5} + 2 }$. In R.H.S , we have $\sf { a\sqrt{5} + b}$.

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$\longrightarrow \sf {\red{ 1}\sqrt{5} + \red{2} \Leftrightarrow \red{a} \sqrt{5} + \red{b }}$

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We get that,

• Value of a ⇒ 1
• Value of b ⇒ 2

More Information :

Some identities :

• (√a)² = a

• √a√b = √ab

• √a/√b = √a/b

• (√a + √b)(√a - √b) = a - b

• (a + √b)(a - √b) = a² - b

• (√a ± √b)² = a ± 2√ab + b

• (√a + √b)(√c + √d) = √ac + √ad + √bc + √bd