MathDB
\sqrt{y^2-8x}+\sqrt{y^2+2x+5} when y=x+2, 1<y<3

Source: Greece JBMO TST 2003 p1

June 18, 2019
algebraComputational

Problem Statement

If point M(x,y)M(x,y) lies on the line with equation y=x+2y=x+2 and 1<y<31<y<3, calculate the value of A=y28x+y2+2x+5A=\sqrt{y^2-8x}+\sqrt{y^2+2x+5}