SOLUTION: On the coordinate plane it is considered the figure(blank) that consist of all points with coordinates AB. Such that system of inequalitiies has no solution. Find the area of the f

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Question 24061: On the coordinate plane it is considered the figure(blank) that consist of all points with coordinates AB. Such that system of inequalitiies has no solution. Find the area of the figure.
x^2+(3-a^2-b^2)x-3(a^2+b^2)<0
and
2x^2+(2a+2b-25)x-25(a+b)>0
Answer by venugopalramana(3286) About Me  (Show Source):
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On the coordinate plane it is considered the figure(blank) that consist of all points with coordinates AB. Such that system of inequalitiies has no solution. Find the area of the figure.
x^2+(3-a^2-b^2)x-3(a^2+b^2)<0
X^2+3X-X(A^2+B^2)-3(A^2+B^2)<0
X(X+3)-(A^2+B^2)(X+3)<0
(X+3)(X-A^2-B^2)<0....THIS HAS NO SOLUTION. THAT IS
(X+3)(X-A^2-B^2)>=0
HENCE X=-3 OR A^2+B^2...TAKING
(X+3)(X-A^2-B^2)=0
and
2x^2+(2a+2b-25)x-25(a+b)>0....DIVIDING WITH 2 THROUGH OUT..
X^2+X(A+B-12.5)-12.5(A+B)>0
X^2+X(A+B)-12.5X-12.5(A+B)>0
X(X+A+B)-12.5(X+A+B)>0
(X+A+B)(X-12.5)>0......THIS HAS NO SOLUTION.THAT IS
(X+A+B)(X-12.5)<=0......
HENCE X=12.5 OR -A-B TAKING
(X+A+B)(X-12.5)=0......SO THE FIGURE HAS AREA BOUNDED BY THE ORDINATES
X=-3,X=A^2+B^2,X=12.5,X=-A-B..AND ABCISSAS Y=0 AND POSITIVE IN THE 1ST.CASE AND Y=0 AND NEGATIVE IN THE 2ND.CASE.
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