Math, asked by anushkanair04, 9 months ago

Root 7 - root 3 + root 5 is​

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Answers

Answered by ahanatarafder06
4

Answer:

\sf{\purple{√7 - √3 + √5 \: is \: an \: Irrational \: Number.}}

Answered by yshfwseem19
2

Answer:

Value of rationalization of \bold{\frac{7+3 \sqrt{5}}{7-3 \sqrt{5}}}7−357+35 is equal to \bold{\frac{(47+21 \sqrt{5 )}}{2}}2(47+215)

Solution:

To start with, rationalization we need to multiply 7+3 \sqrt{5}7+35  with both the numerator and denominator to remove the roots. Multiplying 7+3 \sqrt{5}7+35 we get

\frac{7+3 \sqrt{5}}{7-3 \sqrt{5}} \times \frac{7+3 \sqrt{5}}{7+3 \sqrt{5}}7−357+35×7+357+35

\frac{(7+3 \sqrt{5})^{2}}{(7)^{2}-(3 \sqrt{5})^{2}}=\frac{49+42 \sqrt{5}+45}{49-45}=\frac{94+42 \sqrt{5}}{4}(7)2−(35)2(7+35)2=49−4549+425+45=494+425

(we get the values in (a+b)^{2}(a+b)2  mode in the numerator and (a^{2}-b^{2})(a2−b2) mode in the denominator. Therefore, the primary value after multiplication is  

Taking 2 common in both numerator and denominator we get \frac{94+42 \sqrt{5}}{4}494+425

\bold{\frac{2(47+21 \sqrt{5})}{2.2}=\frac{(47+21 \sqrt{5})}{2}}2.22(47+215)=2(47+215)

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