The value of x^x+2.^x+3.x^2x=where x=2
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Answer:
x
2
+2x,2x+3, & x
2
+3x+8
are the sides of a triangle
then sum of any two side will always be greater than the third side
(x
2
+2x)+(2x+3)>x
2
+3x+8
⇒x>5 ………(1)
(x
2
+2x)+(x
2
+3x+8)>2x+3
2x
2
+3x+5>0
⇒x∈R (∴a>0,D<0)
(2x+3)+(x
2
+3x+8)>x
2
+2x
3x+11>0
x>
3
−11
So these will be sides of a triangle for x>5.
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your answer is x,<5I think it is helpful to you
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