Question 22000
LET A BE THE POSITION OF 50 WATT LAMP AND B BE THE POITION OF THE 60 WATT LAMP
AB=10 M.
LET TONY SIT AT T BETWEEN A AND B.LET AT=X METRES.SO TB=10-X METRES.
INTENSITY IS INVERSELY PROPORTIONAL TO THE DISTANCE FROM THE LAMP.IF WE DENOTE INTENSITY BY I AND DISTACS BY D THEN THIS MEANS
I = K*W/X^2...WHERE K IS A CONSTANT AND W IS THE WATTAGE OF THE LIGHT.(NOTE THAT IF P IS DIRECTLY PROPORTIONAL TO Q THEN P=K*Q,WHERE K IS CONSTANT AND IF P IS INVERSELY PROPORTIONAL TO Q THEN P=K/Q)
NOW THE INTENSITY AT T WHERE TONY SITS IS EQUAL FROM BOTH THE LIGHTS 
INTENSITY FROM  50 WATT LAMP =K*50/X^2
INTENSITY FROM  60 WATT LAMP =K*60/(10-X)^2...THESE ARE EQUAL.HENCE
K*50/X^2=K*60/(10-X)^2....CROSS MULTIPLYING..
50*(10-X)^2=60*X^2
50*[100+X^2-20X]=60*X^2
60*X^2-50*X^2+1000X-5000=0
10X^2+1000X-5000=0
X^2+100X-500=0
USING THE FORMULA FOR ROOTS OF A QUADRATIC EQN.ax^2+bx+c=0
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-100 +- sqrt((-100)^2-4*1*(-500) ))/(2*1) }}}
{{{x = (-100 +- sqrt( 10000+2000))/(2) }}}
{{{x = (-100 +- sqrt( 12000))/(2) }}}
{{{x = (-50 +- 10sqrt( 30)) }}}
{{{x = (-50 +- 10*5.477) }}}
{{{x = (-50 +54.77) }}}
x =4.77