The graphs of y = 2x^2 - 21x + 64 and y = 3x + a, where a is a constant, intersect at exactly one point (x, y). What is the value of x?

Step-by-step solution

1. 1

   Set the two expressions for $y$ equal. At an intersection both curves have the same $y$ for the same $x$:

   $2x^2 - 21x + 64 = 3x + a$

2. 2

   Bring everything to one side. Subtract $3x + a$:

   $2x^2 - 24x + (64 - a) = 0$

   This is a quadratic in $x$ whose constant term still carries the unknown $a$.

3. 3

   Translate "exactly one point" into a discriminant condition. A quadratic has exactly one (repeated) root precisely when its discriminant is zero. With $A = 2$, $B = -24$, $C = 64 - a$:

   $B^2 - 4AC = (-24)^2 - 4(2)(64-a) = 576 - 512 + 8a = 8a + 64 = 0$

4. 4

   Read off $x$ from the repeated root — the shortcut. You do not actually need $a$. When the discriminant vanishes, the double root is $x = -\dfrac{B}{2A}$, the vertex of the parabola $2x^2-24x+(64-a)$:

   $x = \frac{24}{2 \cdot 2} = \frac{24}{4} = 6$

5. 5

   Confirm by finding $a$ anyway. From $8a + 64 = 0$ we get $a = -8$, so the quadratic is $2x^2 - 24x + 72 = 2(x-6)^2 = 0$, giving the double root $x = 6$ ✓.

6. 6

   Check the point lies on both graphs. Parabola: $2(36) - 21(6) + 64 = 72 - 126 + 64 = 10$. Line: $3(6) + (-8) = 10$ ✓. The single intersection point is $(6, 10)$, so $x = 6$.