Question

Find the value of a and b, if (2-root5)/2+root5)=aroot5+b

Answer 1

Hey friend,
$\frac{2 - \sqrt{5} }{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } \\ = ( {2}^{2} ) + ( { \sqrt{5} }^{2} ) - 2 \times 2 \times \sqrt{5} \div ( {2}^{2} ) - ( { \sqrt{5} }^{2} ) \\ = \frac{4 + 5 - 4 \sqrt{5} }{4 - 5} \\ = \frac{9 - 4 \sqrt{5} }{ - 1} \\ = - 9 + 4 \sqrt{5} \\ = 4 \sqrt{5} - 9$
4√5 - 9 = a√5 + b
a = 4
b = -9

Hope my answer would be helpful to you!!!

@Mahak24

Thanks...
☺☺

Answer 2

Answer:

Hey friend,

\begin{gathered} \frac{2 - \sqrt{5} }{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } \\ = ( {2}^{2} ) + ( { \sqrt{5} }^{2} ) - 2 \times 2 \times \sqrt{5} \div ( {2}^{2} ) - ( { \sqrt{5} }^{2} ) \\ = \frac{4 + 5 - 4 \sqrt{5} }{4 - 5} \\ = \frac{9 - 4 \sqrt{5} }{ - 1} \\ = - 9 + 4 \sqrt{5} \\ = 4 \sqrt{5} - 9\end{gathered}

2+

5

2−

5

×

2−

5

2−

5

=(2

2

)+(

5

2

)−2×2×

5

÷(2

2

)−(

5

2

)

=

4−5

4+5−4

5

=

−1

9−4

5

=−9+4

5

=4

5

−9

4√5 - 9 = a√5 + b

a = 4

b = -9

Hope my answer would