Question

the value of x for which the f(x) =25-20x+5x is minimum is

Answer 1

Answer:

x = 5/3

I hope this is helpful for you

Step-by-step explanation:

25-20x+5x=0

25-15x=0

-15x=-25

x= -25/-15

x= 5/3

Answer 2

Step-by-step explanation:

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{x^{2}+y^{2}=98}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \\ \\ \tt: \implies y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \red{\underline \bold{to \: find : }} \\ \tt: \implies {x}^{2} + {y}^{2} = ?$

• According to given question :

$\bold{Rationalising : } \\ \tt: \implies x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ \tt: \implies x = \frac{ (\sqrt{3} + \sqrt{2} )^{2} }{ {( \sqrt{3}})^{2} - { (\sqrt{2} })^{2} } \\ \\ \tt: \implies x = \frac{ {( \sqrt{3} )}^{2} + {( \sqrt{2} )}^{2} + 2 \sqrt{6} }{3 - 2} \\ \\ \tt: \implies x =3 + 2 + 2 \sqrt{6} \\ \\ \tt: \implies x =5 + 2 \sqrt{6} \\ \\ \bold{Similarly : } \\ \tt: \implies y = \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \times \frac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\tt: \implies y = \frac{ (\sqrt{3} - \sqrt{2})^{2} }{ {( \sqrt{3} )}^{2} - { (\sqrt{2} })^{2} } \\ \\ \tt: \implies y= \frac{ { (\sqrt{3} })^{2} + {( \sqrt{2} )}^{2} - 2 \sqrt{6} }{3 - 2 } \\ \\ \tt: \implies y=3 + 2 - 2 \sqrt{6} \\ \\ \tt: \implies y =5 - 2 \sqrt{6} \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies {x}^{2} + {y}^{2} =(x + y)^{2} - 2xy \\ \\ \tt: \implies {x}^{2} + {y}^{2} =(5 + 2 \sqrt{6} + 5 - 2 \sqrt{6} )^{2} - 2(5 + 2\sqrt{6} )(5 - 2 \sqrt{6} ) \\ \\ \tt: \implies {x}^{2} + {y}^{2} = {10}^{2} - 2( {5}^{2} - {(2 \sqrt{6}) }^{2} \\ \\ \tt: \implies {x}^{2} + {y}^{2} =100 - 2(25 - 24) \\ \\ \tt: \implies {x}^{2} + {y}^{2} =100 - 2 \\ \\ \green{\tt: \implies {x}^{2} + {y}^{2} =98}$