Algebric problem

[Richard Hachel]

One commenter posed the problem:

sqrt(3x+7)+sqrt(x+2)=1

On the French forum, mathematicians had some problems with this.

However, the solution is very simple, and requires three lines of
calculations on a piece of paper.

R.H.

[Mike Terry]

Post by Richard Hachel
sqrt(3x+7)+sqrt(x+2)=1
On the French forum, mathematicians had some problems with this.
However, the solution is very simple, and requires three lines of calculations on a piece of paper.
R.H.

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Square both sides + rearrange giving
..linear fn of x.. = sqrt(..quadratic fn of x..)
Square both sides giving
..quadratic fn of x.. = ..another quadratic fn of x..
Solve quadratic giving 2 possibilities for x.
Check back with original equation (since squaring may introduce extra solutions)

My working gives the quadratic x^2+3x+2 = 0 solutions x = -1 or x = -2.
Checking with original equation: x=-1 fails. x=-2 works, so only solution is x=-2.

Mike.

[Jim Burns]

Post by Richard Hachel
sqrt(3x+7)+sqrt(x+2)=1
On the French forum,
mathematicians had some problems with this.
However,
the solution is very simple,
and requires three lines of calculations
on a piece of paper.

Is the problem that
some of the possible solutions presented by
three lines of calculations
do not produce equality in the original equation?

Squaring can introduce more possible solutions.

for each x such that √(3x+7)+√(x+2)=1
also √(3x+7)=1-√(x+2)

for each x such that √(3x+7)=1-√(x+2)
also 3x+7=1-2√(x+2)+x+2
but also
for each x such that -√(3x+7)=1-√(x+2)
also 3x+7=1-2√(x+2)+x+2

for each x such that
√(3x+7)=1-√(x+2) or
-√(3x+7)=1-√(x+2)
also
3x+7=1-2√(x+2)+x+2 and
x+2=-√(x+2)

for each x such that
√(3x+7)=1-√(x+2) or
-√(3x+7)=1-√(x+2) or
x+2=-√(x+2) or
x+2=√(x+2)
also
(x+2)²=x+2
(x+1)(x+2)=0
x=-1 or x=-2

Only -1 and -2 _can be_ a solution.

√(3⋅-1+7)+√(-1+2)=√4+√1≠1

√(3⋅-2+7)+√(-2+2)=√1+√0=1

Only -2 _is_ a solution.

Post by Richard Hachel
R.H.

[Ross Finlayson]

Post by Jim Burns

Post by Richard Hachel
sqrt(3x+7)+sqrt(x+2)=1
On the French forum,
mathematicians had some problems with this.
However,
the solution is very simple,
and requires three lines of calculations
on a piece of paper.

Is the problem that
some of the possible solutions presented by
three lines of calculations
do not produce equality in the original equation?
Squaring can introduce more possible solutions.
for each x such that √(3x+7)+√(x+2)=1
also √(3x+7)=1-√(x+2)
for each x such that √(3x+7)=1-√(x+2)
also 3x+7=1-2√(x+2)+x+2
but also
for each x such that -√(3x+7)=1-√(x+2)
also 3x+7=1-2√(x+2)+x+2
for each x such that
√(3x+7)=1-√(x+2) or
-√(3x+7)=1-√(x+2)
also
3x+7=1-2√(x+2)+x+2 and
x+2=-√(x+2)
for each x such that
√(3x+7)=1-√(x+2) or
-√(3x+7)=1-√(x+2) or
x+2=-√(x+2) or
x+2=√(x+2)
also
(x+2)²=x+2
(x+1)(x+2)=0
x=-1 or x=-2
Only -1 and -2 _can be_ a solution.
√(3⋅-1+7)+√(-1+2)=√4+√1≠1
√(3⋅-2+7)+√(-2+2)=√1+√0=1
Only -2 _is_ a solution.

Post by Richard Hachel
R.H.

y = 3x+7
z = x+2

root y = 1 - root z && root z = 1 - root y -> 1 >= y, z >= 0

y <= 1 -> 3x+7 <= 1 -> 3x <= 8 -> x <= 8/3
z <= 1 -> x+2 <= 1 -> x < 3 -> x <= 3
y >= 0 -> 3x+7 >= 0 -> 3x >= -7 -> x >= -7/3
z >= 0 -> x+2 >= 0 -> x > -2
-> -2 <= x <= 8/3

[Python]

Post by Richard Hachel
sqrt(3x+7)+sqrt(x+2)=1
On the French forum, mathematicians had some problems with this.

No "mathematician", and almost none of the group usual participants
has any issues. Only the initial poster (obviously), *you* (making
a lot of silly mistakes and comments) and Jacques Lavau (as usual)
had issues.

Post by Richard Hachel
However, the solution is very simple, and requires three lines of
calculations on a piece of paper.

Why are you lying and making a fool of yourself in such an idiotic
way, M.D. Richard "Hachel" Lengrand?

[Richard Hachel]

Post by Richard Hachel
sqrt(3x+7)+sqrt(x+2)=1
However, the solution is very simple, and requires three lines of
calculations on a piece of paper.

Why are you lying ?

Because I like to lie.

To lie is so wonderfull, so orgasmic.

Tu devrais essayer, si tu savais comme c'est bon.

R.H.