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

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) (Show Source): You can put this solution on YOUR website!
SEE MY COMMENTS BELOW
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.
CAN YOU PLEASE CHECK BACK AND GIVE ME A SCAN OR PRINT OF THE PROBLEM AS GIVEN OR GOT BY YOU TO COMPLETE THE SOLUTION FROM HERE AS I HAVE GOT SOME DOUBT ON THE WRITE UP GIVEN BY YOU
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