(2+sqrt{5})^{2}=?+4 sqrt{5} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To solve the expression $(2+\sqrt{5})^{2}$,we use the algebraic identity $(a+b)^{2} = a^{2} + b^{2} + 2ab$.Here,$a = 2$ and $b = \sqrt{5}$.$(2+\sq

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) To solve the expression $(2+\sqrt{5})^{2}$,we use the algebraic identity $(a+b)^{2} = a^{2} + b^{2} + 2ab$.Here,$a = 2$ and $b = \sqrt{5}$.$(2+\sqrt{5})^{2} = (2)^{2} + (\sqrt{5})^{2} + 2(2)(\sqrt{5})$.$= 4 + 5 + 4\sqrt{5}$.$= 9 + 4\sqrt{5}$.Comparing this with the given equation $? + 4\sqrt{5}$,we get $9 + 4\sqrt{5} = ? + 4\sqrt{5}$.Therefore,$? = 9$.

$(2+\sqrt{5})^{2}=?+4 \sqrt{5}$

Similar Questions

The sum of the first $45$ natural numbers is:

$A$ number when divided by $221$ gives a remainder $43$. What remainder will be obtained by dividing the same number by $17$?

$63 + 371 \div 7 = ?$

The sum and product of two numbers are $12$ and $35$ respectively. The sum of their reciprocals will be

$n$ is a whole number which when divided by $4$ gives the remainder $3$. The remainder when $2n$ is divided by $4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module