For each question,select the proper option fr | vedclass

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

Question

(B) To determine the nature of the number $(\sqrt{5}+3)^{2}$,we expand it using the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$.Here,$a = \sq

Vedclass · Accepted Answer

irrational

Vedclass · Answer

(B) To determine the nature of the number $(\sqrt{5}+3)^{2}$,we expand it using the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$.Here,$a = \sqrt{5}$ and $b = 3$.$(\sqrt{5}+3)^{2} = (\sqrt{5})^{2} + 2(\sqrt{5})(3) + (3)^{2}$$= 5 + 6\sqrt{5} + 9$$= 14 + 6\sqrt{5}$.Since $\sqrt{5}$ is an irrational number,$6\sqrt{5}$ is also irrational. The sum of a rational number $(14)$ and an irrational number $(6\sqrt{5})$ is always an irrational number.Therefore,$(\sqrt{5}+3)^{2}$ is an irrational number.

For each question,select the proper option from the four options given to make the statement true: $(\sqrt{5}+3)^{2}$ is a / an $\ldots \ldots \ldots$ number.

Similar Questions

State whether the following statement is true or false:
$\sqrt{49}$ is an irrational number.

Prove that $(1^{3}+2^{3}+3^{3}+4^{3}+5^{3})^{\frac{1}{2}} = (1^{3}+2^{3}+3^{3}+4^{3})^{\frac{1}{2}} + (5^{3})^{\frac{1}{3}}$.

Between two rational numbers,

Simplify the following:
$\frac{2 \sqrt{3}}{3} - \frac{\sqrt{3}}{6}$

Prove that $\frac{x^{a(b-c)}}{x^{b(a-c)}} \div \left(\frac{x^{b}}{x^{a}}\right)^{c} = 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module