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

Step-by-step solution

1. 1

   Turn the intersection condition into one equation. A point lies on both graphs when the two expressions for $y$ agree, so set them equal:

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

   Every intersection point corresponds to a solution of this equation, and vice versa.

2. 2

   Move everything to one side. Collecting terms gives a quadratic in standard form:

   $2x^2-21x-3x+64-a=0$

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

   The unknown constant $a$ sits only in the constant term — that observation is what makes the problem short.

3. 3

   Translate 'exactly one intersection point' into 'exactly one root'. A quadratic has exactly one real solution precisely when its discriminant vanishes:

   $\Delta=b^2-4ac=(-24)^2-4(2)(64-a)=0$

   Geometrically this is the statement that the line is tangent to the parabola instead of cutting it twice or missing it.

4. 4

   Use the shortcut: a double root sits at the vertex of the quadratic. When $\Delta=0$ the two roots coincide at their average, $x=-\dfrac{b}{2a}$. Here $b=-24$ and the leading coefficient is $2$, so

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

   Notice that $a$ never entered this computation: it only shifts the constant term, which does not move the axis of symmetry. That is why the question can ask for $x$ without telling you $a$.

5. 5

   Find $a$ if you want to see the tangency explicitly. Solving $\Delta=0$:

   $576-8(64-a)=0\;\Longrightarrow\;576-512+8a=0\;\Longrightarrow\;a=-8$

   The quadratic becomes $2x^2-24x+72=2(x-6)^2=0$, confirming the repeated root $x=6$.

6. 6

   Verify the single intersection point. With $x=6$ the parabola gives $2(36)-21(6)+64=72-126+64=10$, and the line gives $3(6)-8=10$. Both curves pass through $(6,10)$ and, since the discriminant is zero, they meet nowhere else:

   $\boxed{x=6}$