Math, asked by deepak436023, 1 year ago

under root x minus 4 + under root 6 minus x is greater than zero​

Answers

Answered by jass9584
0

 \sqrt{x - 4}  +  \sqrt{6 - x}  > 0
 \sqrt{x - (2 \times 2)}  +  \sqrt{(2 \times 3) - x}  > 0
Taking root 2 as common....

 \sqrt{2} ( \sqrt{x - 2}  +  \sqrt{3 -  x} ) > 0
 \sqrt{x - 2}  +  \sqrt{3 - x}  >  \frac{0}{ \sqrt{2} }
Now by squaring both sides.....

 {(\sqrt{x - 2}  +  \sqrt{3 - x}) }^{2}  >  {0}^{2}
x - 2 + 3 - x > 0
 - 2 + 3 > 0
1 > 0

as 1 is greater than 0 so

HENCE PROVED.........
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