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Greece Contests
Greece JBMO TST
2003 Greece JBMO TST
1
1
Part of
2003 Greece JBMO TST
Problems
(1)
\sqrt{y^2-8x}+\sqrt{y^2+2x+5} when y=x+2, 1<y<3
Source: Greece JBMO TST 2003 p1
6/18/2019
If point
M
(
x
,
y
)
M(x,y)
M
(
x
,
y
)
lies on the line with equation
y
=
x
+
2
y=x+2
y
=
x
+
2
and
1
<
y
<
3
1<y<3
1
<
y
<
3
, calculate the value of
A
=
y
2
−
8
x
+
y
2
+
2
x
+
5
A=\sqrt{y^2-8x}+\sqrt{y^2+2x+5}
A
=
y
2
−
8
x
+
y
2
+
2
x
+
5
algebra
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