Question

If (√5+√2)^2=a+b√10 find a and b

Answer 1

Using the identity, (a + b)² = a² + b² + 2ab, we get

(√5 + √2)²

=> (√5)² + (√2)² + 2(√5)(√2)

=> 5 + 2 + 2√10

=> 7 + 2√10

Now given that,

(√5 + √2)² = a + b√10

but we have derived that,

(√5 + √2)² = 7 + 2√10

=> 7 + 2√10 = a + b√10

Now √10 has coefficient b in RHS and coefficient 2 in LHS. Since √10 is common in them we can say,

b = 2
And we are left with 7 and a

=> a = 7

This method is called as comparison method.


Hope it helps dear friend ☺️

Answer 2

$\text{Answer :}$

$= > {( \sqrt{5} + \sqrt{2}) }^{2} \\ \\ = > {( \sqrt{5} )}^{2} + {( \sqrt{2} )}^{2} + 2 \times \sqrt{5} \times \sqrt{2} \\ \\ = > (\sqrt{5 \times 5}) + (\sqrt{2 \times 2}) + 2 \times (\sqrt{5 \times 2}) \\ \\ = > 5 + 2 + 2 \sqrt{10} \\ \\ = > 7 + 2 \sqrt{10}$

$\textbf{Identity used : }$

$= > {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Let's find the value of $\underline\color\textbf{a}$ and $\underline\color\textbf{b}$

=> 7 + 2√10 = a + b√10

√10 is the common term present in both LHS and RHS, hence, √10 won't be included in the values.

$\textbf{Hence, }$

$\underline\textbf{= 7 = a}$

$\underline\textbf{= 2 = b}$

$\text{Tysm for the question }$^‿^