Question

If 1/(√5-2)=a√5+b then what is value of a and b

Answer 1

$\bf\large{\underline{Question:-}}$

If 1/(√5-2)=a√5+b then what is value of a and b

$\bf\large{\underline{To\:find:-}}$

• Value of a and b

Identity used

$\tt{\fbox{\red{\underline{a^2-b^2=(a+b)(a-b)}}}}$

$\bf\large{\underline{Solution:-}}$

$\tt→ \frac{1}{\sqrt5 -2}×\frac{\sqrt5+2}{\sqrt5+2}=a\sqrt5+b\\\tt→ \frac{\sqrt5+2}{\sqrt5^2-2^2}=a\sqrt5+b\\\tt→ \frac{\sqrt5+2}{5-4}=a\sqrt5+b\\\tt→ \sqrt5+2=a\sqrt5+b\\\tt→ a\sqrt5=\sqrt5\\\tt→ a= 1\\\tt→ b=2$

Hence,

★ a=1

★ b=2

Answer 2

Answer:

The answer is, a= 1 & b=2

Step-by-step explanation:

We are given,

$1/\sqrt{5} -2=a\sqrt{5} +2$

By rationalising the denominator,

$\frac{1}{\sqrt{5}-2 } =\frac{1}{\sqrt{5}-2 } *\frac{\sqrt{5}+2 }{\sqrt{5}+2 }$

$=\frac{\sqrt{5} +2}{(\sqrt{5})^{2}-2^{2} } =\frac{\sqrt{5}+2 }{1}$

$=\sqrt{5} +2=a\sqrt{5} +b$

Now by comparison,

we can easily see that,

a=1 and b=2