( x + 1 + 1 ) ( x + 1 6 + 4 ) − ( x + 4 + 2 ) ( x + 9 + 3 ) = 0
How many real value(s) of x satisfy the above equation?
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You have only shown that there is no solution in a finite interval, can you show that there is no solution throughout the whole domain?
And, can you solve this without graphing?
Is there any other way to solve this .....?
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There sure is, but it would involve tedious algebra and the application of many formulas...why do that when there is such an easy way to solve it! :)
Yes, just look at the RHS and LHS. Two irrational quantities can be made equal only by making the rational and irrational parts equal individually, like in the case of complex numbers. Also, from the expression you can say the irrational parts will never be equal. Hence, no solution.
guys give me credit plzzz i was first to solve here.
-1- ( (x+1)^(1/2) + 1 )( (x+16)^(1/2) + 4 ) - ( (x+4)^(1/2) + 2 )( (x+9)^(1/2) + 3 ) = 0
A= ( (x + (a^2) )^(1/2) +a ) = x/[( (x + (a^2) )^(1/2) - a )] taking x common and manipulating eqn -1- little bit
-2- ( (x+1)^(1/2) - 1 )( (x+16)^(1/2) - 4 ) - ( (x+4)^(1/2) - 2 )( (x+9)^(1/2) - 3 ) = 0
-3- opening brackets for eqn 1 and 2 and adding both of them we get:- [(x+1)(x+16)] + 4 -[(x+4)(x+9)]-6=0
-4- after solving we may get x==0 rejecting that solution we have no other fundamental soln..
btw sry i pressed the wrong answer by mistake
Man I didn't understand it can you plz be a bit elaborate
The graph of the
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0n the calculator shows that y=0 is the asymptote of the curve. So y=0 is never reached. Implies no solution. the window was set to x= 12 * 7 and y=-.001.
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( x + 1 + 1 ) ( x + 1 6 + 4 ) − ( x + 4 + 2 ) ( x + 9 + 3 ) = 0
→ ( x + 1 + 1 ) ( x + 1 6 + 4 ) = ( x + 4 + 2 ) ( x + 9 + 3 )
So, the values of x that will satisfy the above equation will the points that coincide when the graphs of L H S and R H S are drawn.
The graphs are found to be parallel to each other
Hence there do not exist any solutions for x