Math, asked by funnylol5427, 16 days ago

# the expression x^2-3x+4/x^2+3x+4 lies between

0

Step-by-step explanation:

Let y =x2+3x+4x2−3x+4

Let y =x2+3x+4x2−3x+4(y−1)x2+3(y+1)x+4(y−1)=0

Let y =x2+3x+4x2−3x+4(y−1)x2+3(y+1)x+4(y−1)=0Since x is real , D≥0

Let y =x2+3x+4x2−3x+4(y−1)x2+3(y+1)x+4(y−1)=0Since x is real , D≥09(y2+2y+1)−16(y2−2y+1)≥0

Let y =x2+3x+4x2−3x+4(y−1)x2+3(y+1)x+4(y−1)=0Since x is real , D≥09(y2+2y+1)−16(y2−2y+1)≥0 −7(y2+1)+50y≥0

Let y =x2+3x+4x2−3x+4(y−1)x2+3(y+1)x+4(y−1)=0Since x is real , D≥09(y2+2y+1)−16(y2−2y+1)≥0 −7(y2+1)+50y≥07y2−50y+7≤0

Let y =x2+3x+4x2−3x+4(y−1)x2+3(y+1)x+4(y−1)=0Since x is real , D≥09(y2+2y+1)−16(y2−2y+1)≥0 −7(y2+1)+50y≥07y2−50y+7≤0(7y−1)(y−7)≤0

Let y =x2+3x+4x2−3x+4(y−1)x2+3(y+1)x+4(y−1)=0Since x is real , D≥09(y2+2y+1)−16(y2−2y+1)≥0 −7(y2+1)+50y≥07y2−50y+7≤0(7y−1)(y−7)≤0y lies in [71,7]

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