2Number 2ExplanationIf the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 has no real roots, then the graph of the function y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c is tangent to the line y=−xy = -xy=−x when....b<−12b < -\frac{1}{2}b<−21−12<b<0-\frac{1}{2} < b < 0−21<b<0b>−12b > -\frac{1}{2}b>−210<b<120 < b < \frac{1}{2}0<b<21b>0b > 0b>0ExplanationThe condition for a quadratic equation to have no real roots is D<0D < 0D<0 b2−4ac<0...(1)b^2 - 4ac < 0 \quad ...(1)b2−4ac<0...(1) Because the line is tangent to the curve, the relationship between them applies ax2+bx+c=−xax^2 + bx + c = -xax2+bx+c=−xax2+bx+x+c=0ax^2 + bx + x + c = 0ax2+bx+x+c=0ax2+(b+1)x+c=0ax^2 + (b + 1)x + c = 0ax2+(b+1)x+c=0 The condition for tangency is D=0D = 0D=0 (b+1)2−4ac=0(b + 1)^2 - 4ac = 0(b+1)2−4ac=0 4ac=(b+1)2...(2)4ac = (b + 1)^2 \quad ...(2)4ac=(b+1)2...(2) Substitute equation (2)(2)(2) into (1)(1)(1) b2−4ac<0b^2 - 4ac < 0b2−4ac<0b2−(b+1)2<0b^2 - (b + 1)^2 < 0b2−(b+1)2<0b2−(b2+2b+1)<0b^2 - (b^2 + 2b + 1) < 0b2−(b2+2b+1)<0b2−b2−2b−1<0b^2 - b^2 - 2b - 1 < 0b2−b2−2b−1<0−2b−1<0-2b - 1 < 0−2b−1<0−2b<1-2b < 1−2b<1b>−12b > -\frac{1}{2}b>−21
