To determine which line will have no solution with the parabola [tex]\( y - x + 2 = x^2 \)[/tex], we need to examine the points of intersection between the given parabola and a generic straight line given by [tex]\( y = mx + c \)[/tex].
We start by substituting [tex]\( y = mx + c \)[/tex] into the parabola equation:
[tex]\[ (mx + c) - x + 2 = x^2 \][/tex]
Simplifying, we obtain:
[tex]\[ mx + c - x + 2 = x^2 \][/tex]
which simplifies to:
[tex]\[ x^2 - (m+1)x + (c + 2) = 0 \][/tex]
This is a quadratic equation in [tex]\(x\)[/tex]:
[tex]\[ x^2 - (m+1)x + (c + 2) = 0 \][/tex]
For the line to have no intersection with the parabola, this quadratic equation must have no real solutions. For a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the condition for having no real solutions is that the discriminant must be negative. The discriminant [tex]\(\Delta\)[/tex] of the quadratic equation [tex]\( x^2 - (m+1)x + (c + 2) = 0 \)[/tex] is:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -(m+1)\)[/tex], and [tex]\(c = c + 2\)[/tex]. Substituting these values into the discriminant formula, we get:
[tex]\[ \Delta = (-(m+1))^2 - 4 \cdot 1 \cdot (c + 2) \][/tex]
[tex]\[ \Delta = (m+1)^2 - 4(c + 2) \][/tex]
For there to be no real solutions:
[tex]\[ \Delta < 0 \][/tex]
So:
[tex]\[ (m+1)^2 - 4(c + 2) < 0 \][/tex]
[tex]\[ (m+1)^2 < 4(c + 2) \][/tex]
[tex]\[ (m+1)^2 < 4c + 8 \][/tex]
Solving for [tex]\(c\)[/tex]:
[tex]\[ (m+1)^2 < 4c + 8 \][/tex]
[tex]\[ 4c + 8 > (m+1)^2 \][/tex]
[tex]\[ 4c > (m+1)^2 - 8 \][/tex]
[tex]\[ c > \frac{(m+1)^2 - 8}{4} \][/tex]
Therefore, the line [tex]\( y = mx + c \)[/tex] will have no intersection with the parabola [tex]\( y - x + 2 = x^2 \)[/tex] if the constant [tex]\( c \)[/tex] satisfies:
[tex]\[ c > \frac{(m+1)^2 - 8}{4} \][/tex]
This inequality determines the values of [tex]\( c \)[/tex] for which the given line does not intersect with the parabola.
