Question 84·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

After setting the two expressions for $y$ equal, expand $(x - 3)^2$ and collect like terms to form a standard quadratic equation $ax^2 + bx + k = 0$.

For the quadratic you formed, determine which condition on the discriminant $b^2 - 4ac$ guarantees one or zero real solutions for $x$.

After solving the discriminant inequality for $c$, use the fact that $c$ must be an integer to identify the smallest allowable integer.

In Desmos, enter $y = (x + 10)/2$ for the line. Then enter $y = (x - 3)^2 + c$ and create a slider for $c$. The slider moves the parabola up and down.

Move the $c$ slider and observe the number of intersection points between the line and the parabola. Identify the boundary value of $c$ at which the line touches the parabola at exactly one point; above that value, there are no intersections.

Use the boundary value shown in Desmos to determine the smallest integer value of $c$ that is at or above the cutoff for at most one intersection.

Rewrite the linear equation in slope-intercept form:

Both equations now express $y$:

Set them equal to find the $x$-values where the graphs intersect:

Expand $(x - 3)^2$ and move everything to one side:

Substitute this into the equation:

Bring all terms to the left side:

Combine like terms:

To avoid decimals, multiply the entire equation by $2$:

For a quadratic equation of the form $ax^2 + bx + k = 0$, the discriminant is

There are:

• Two real solutions if $D > 0$,
• One real solution, a repeated root, if $D = 0$,
• No real solutions if $D < 0$.

“At most one real solution” means $D \le 0$.

In the equation $2x^2 - 13x + (8 + 2c) = 0$, the coefficients are $a = 2$, $b = -13$, and $k = 8 + 2c$. Thus,

Simplify:

Now impose $D \le 0$:

Solve the inequality:

Add $16c$ to both sides:

Then divide by $16$:

Because $c$ is an integer, the smallest possible value satisfying this inequality is $7$.