Mixing
Inequality with square roots where solution found with discriminant is not valid
$begingroup$
I have :
$3 + sqrt{x-1} > sqrt{2x}$
when doing basic algebraic operations
I will find that :
$ 0 > x^{2} -52x + 100 $
I will use $b^2-4ac$ formula
to eventually find out that there are 2 roots $x_1 = 50 $ and $ x_2 = 2$
However I will find out when testing these two results that $x_2$ is not the solution.
I also know that $ x > 1$
The result should be $ x[1,50) $ but I can not understand how can I interpolate this result from results above.
Is there some fault in my reasoning to find solution with discriminant?
algebra-precalculus inequality radicals discriminant
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
asked Jan 21 at 10:51
cris14cris14
1338
$endgroup$
add a comment |
$begingroup$
I have :
$3 + sqrt{x-1} > sqrt{2x}$
when doing basic algebraic operations
I will find that :
$ 0 > x^{2} -52x + 100 $
I will use $b^2-4ac$ formula
to eventually find out that there are 2 roots $x_1 = 50 $ and $ x_2 = 2$
However I will find out when testing these two results that $x_2$ is not the solution.
I also know that $ x > 1$
The result should be $ x[1,50) $ but I can not understand how can I interpolate this result from results above.
Is there some fault in my reasoning to find solution with discriminant?
algebra-precalculus inequality radicals discriminant
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
asked Jan 21 at 10:51
cris14cris14
1338
$endgroup$
4
$begingroup$
Squaring always introduces extra solutions (to be checked).
$endgroup$
– Claude Leibovici
Jan 21 at 10:55
add a comment |
$begingroup$
I have :
$3 + sqrt{x-1} > sqrt{2x}$
when doing basic algebraic operations
I will find that :
$ 0 > x^{2} -52x + 100 $
I will use $b^2-4ac$ formula
to eventually find out that there are 2 roots $x_1 = 50 $ and $ x_2 = 2$
However I will find out when testing these two results that $x_2$ is not the solution.
I also know that $ x > 1$
The result should be $ x[1,50) $ but I can not understand how can I interpolate this result from results above.
Is there some fault in my reasoning to find solution with discriminant?
algebra-precalculus inequality radicals discriminant
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
asked Jan 21 at 10:51
cris14cris14
1338
$endgroup$
I have :
$3 + sqrt{x-1} > sqrt{2x}$
when doing basic algebraic operations
I will find that :
$ 0 > x^{2} -52x + 100 $
I will use $b^2-4ac$ formula
to eventually find out that there are 2 roots $x_1 = 50 $ and $ x_2 = 2$
However I will find out when testing these two results that $x_2$ is not the solution.
I also know that $ x > 1$
The result should be $ x[1,50) $ but I can not understand how can I interpolate this result from results above.
Is there some fault in my reasoning to find solution with discriminant?
algebra-precalculus inequality radicals discriminant
algebra-precalculus inequality radicals discriminant
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
asked Jan 21 at 10:51
cris14cris14
1338
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
asked Jan 21 at 10:51
cris14cris14
1338
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
edited Jan 21 at 11:03
Michael Rozenberg
107k1894198
107k1894198
asked Jan 21 at 10:51
cris14cris14
1338
asked Jan 21 at 10:51
cris14cris14
1338
asked Jan 21 at 10:51
cris14cris14
1338
1338
4
$begingroup$
Squaring always introduces extra solutions (to be checked).
$endgroup$
– Claude Leibovici
Jan 21 at 10:55
add a comment |
4
$begingroup$
Squaring always introduces extra solutions (to be checked).
$endgroup$
– Claude Leibovici
Jan 21 at 10:55
4
4
$begingroup$
Squaring always introduces extra solutions (to be checked).
$endgroup$
– Claude Leibovici
Jan 21 at 10:55
$begingroup$
Squaring always introduces extra solutions (to be checked).
$endgroup$
– Claude Leibovici
Jan 21 at 10:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The domain gives $xgeq1$ and we have
$$9+6sqrt{x-1}+x-1>2x$$ or
$$6sqrt{x-1}>x-8.$$
Now, for $xleq8$ it's true, but for $x>8$ we obtain
$$36(x-1)>(x-8)^2$$ or
$$x^2-52x+100<0$$ or
$$(x-2)(x-50)<0$$ or $$2<x<50,$$ which with $x>8$ gives $$8<x<50$$ and we got the answer:
$$[1,50).$$
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The domain gives $xgeq1$ and we have
$$9+6sqrt{x-1}+x-1>2x$$ or
$$6sqrt{x-1}>x-8.$$
Now, for $xleq8$ it's true, but for $x>8$ we obtain
$$36(x-1)>(x-8)^2$$ or
$$x^2-52x+100<0$$ or
$$(x-2)(x-50)<0$$ or $$2<x<50,$$ which with $x>8$ gives $$8<x<50$$ and we got the answer:
$$[1,50).$$
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
The domain gives $xgeq1$ and we have
$$9+6sqrt{x-1}+x-1>2x$$ or
$$6sqrt{x-1}>x-8.$$
Now, for $xleq8$ it's true, but for $x>8$ we obtain
$$36(x-1)>(x-8)^2$$ or
$$x^2-52x+100<0$$ or
$$(x-2)(x-50)<0$$ or $$2<x<50,$$ which with $x>8$ gives $$8<x<50$$ and we got the answer:
$$[1,50).$$
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
add a comment |
$begingroup$
The domain gives $xgeq1$ and we have
$$9+6sqrt{x-1}+x-1>2x$$ or
$$6sqrt{x-1}>x-8.$$
Now, for $xleq8$ it's true, but for $x>8$ we obtain
$$36(x-1)>(x-8)^2$$ or
$$x^2-52x+100<0$$ or
$$(x-2)(x-50)<0$$ or $$2<x<50,$$ which with $x>8$ gives $$8<x<50$$ and we got the answer:
$$[1,50).$$
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
$endgroup$
The domain gives $xgeq1$ and we have
$$9+6sqrt{x-1}+x-1>2x$$ or
$$6sqrt{x-1}>x-8.$$
Now, for $xleq8$ it's true, but for $x>8$ we obtain
$$36(x-1)>(x-8)^2$$ or
$$x^2-52x+100<0$$ or
$$(x-2)(x-50)<0$$ or $$2<x<50,$$ which with $x>8$ gives $$8<x<50$$ and we got the answer:
$$[1,50).$$
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
answered Jan 21 at 10:59
Michael RozenbergMichael Rozenberg
107k1894198
107k1894198
add a comment |
add a comment |
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4
$begingroup$
Squaring always introduces extra solutions (to be checked).
$endgroup$
– Claude Leibovici
Jan 21 at 10:55