Question

if (2-√5)+(3+2√5)-(4+√5)/4 = p+q√5, then value of p and q are​

Answer 1

Step-by-step explanation:

$\tt{ \implies \: \frac{ (2- \sqrt{ 5 } )+(3+2 \sqrt{ 5 } )-(4+ \sqrt{ 5 } ) }{ 4 } =p+q \sqrt{ 5 }}$

$\tt{ \implies \:2-\sqrt{5}+3+2\sqrt{5}-\left(4+\sqrt{5}\right)=4p+4q\sqrt{5} }$

$\tt{ \implies \:5-\sqrt{5}+2\sqrt{5}-\left(4+\sqrt{5}\right)=4p+4q\sqrt{5} }$

$\tt{ \implies \:5+\sqrt{5}-\left(4+\sqrt{5}\right)=4p+4q\sqrt{5} }$

$\tt{ \implies5+\sqrt{5}-4-\sqrt{5}=4p+4q\sqrt{5} }$

$\tt{ \implies \: 1+\sqrt{5}-\sqrt{5}=4p+4q\sqrt{5} }$

$\tt{ \implies \: 1=4p+4q\sqrt{5} }$

$\tt{ \implies \: 4p=1-4q\sqrt{5} }$

$\tt{ \implies \:\frac{4p}{4}=\frac{-4\sqrt{5}q+1}{4} }$

$\tt{ \implies \:p=-\sqrt{5}q+\frac{1}{4} }$

Answer:

Step-by-step explanation:

hope it help you.

thanks

Answer 2

Answer:

given equation : $\frac{(2-\sqrt{5}) + (3 + 2\sqrt{5}) - (4 + \sqrt{5} ) }{4} = p + q\sqrt{5}$

Step-by-step explanation:

now , on  multiplying by 4 and solving we get ;

$= 1 -\sqrt{5} + 2\sqrt{5} - \sqrt{5} = 4p +4q\sqrt{5}$

$1 = 4p +4q\sqrt{5}$

$4p = 1 -4q\sqrt{5}$

$p = \frac{1}{4} - q\sqrt{5}$

$q = \frac{\frac{1}{4} - p }{\sqrt{5} }$