Once you have the quadratic in $x$, recall that the discriminant $B^2 - 4AC$ tells you how many real solutions it has. What inequality should that discriminant satisfy if you want no real solutions?

Identify $A$, $B$, and $C$ in your quadratic, write the discriminant using $a$ and $b$, set it less than 0, and then solve this inequality for $b$.

For each integer $a$ from 0 to 6, use your inequality to find all integer $b$ between 0 and 10 that work. Count how many $b$ values you get for each $a$, then add those counts.

From the algebra, you should have the condition for no real solutions as $b < \dfrac{20 - (a+1)^2}{4}$. In Desmos, let the horizontal axis represent $a$ and the vertical axis represent $b$, and type the function:

This curve is the boundary separating where the discriminant is negative from where it is not.

In Desmos, type the inequalities

• 0 <= x <= 6
• 0 <= y <= 10

to shade the rectangle that represents all allowed integer pairs $(a,b)$.

Now type the inequality

This shades the region of $(a,b)$ for which the discriminant is negative (no real solutions). Look only inside the rectangle $0 \le x \le 6$, $0 \le y \le 10$, and count how many grid intersection points with integer coordinates $(x,y)$ lie in the shaded region. That count is the number of ordered pairs $(a,b)$ that satisfy the problem.

Because both expressions equal $y$, set them equal to each other:

Move all terms to one side to get a standard quadratic in $x$:

This equation must have no real solutions in $x$ for the system to have no real solutions.

For a quadratic equation

the discriminant is $B^2 - 4AC$:

• If $B^2 - 4AC > 0$, there are 2 real solutions.
• If $B^2 - 4AC = 0$, there is 1 real solution (a tangent).
• If $B^2 - 4AC < 0$, there are no real solutions.

In our quadratic $x^2 + (a+1)x + (5 - b) = 0$:

• $A = 1$
• $B = a + 1$
• $C = 5 - b$

So the discriminant is

We want no real solutions, so we need

Start from

Distribute the $4$:

Move terms so that $b$ is on one side:

and divide by $4$:

We must also remember the given conditions:

• $a$ is an integer with $0 \le a \le 6$.
• $b$ is an integer with $0 \le b \le 10$.

So for each integer $a$ from $0$ to $6$, $b$ must be an integer between $0$ and $10$ that also satisfies this inequality.

Compute the right side of

for each integer $a$ from $0$ to $6$.

• $a = 0$:

With $0 \le b \le 10$ and integer $b$, we can have $b = 0,1,2,3,4$ → 5 values.

• $a = 1$:

So $b$ can be $0,1,2,3$ → 4 values.

• $a = 2$:

So $b$ can be $0,1,2$ → 3 values.

• $a = 3$:

So $b$ can only be $0$ → 1 value.

• $a = 4$:

But $b \ge 0$, so no values of $b$ work.

• $a = 5$:

Again, impossible with $b \ge 0$ → 0 values.

• $a = 6$:

Also impossible with $b \ge 0$ → 0 values.

So the numbers of valid $b$ values for $a = 0,1,2,3,4,5,6$ are respectively

Each allowed $(a,b)$ pair corresponds to a choice of $a$ (from $0$ to $6$) and a $b$ that satisfies the inequality for that $a$.

From the previous step, we have:

• 5 pairs when $a = 0$
• 4 pairs when $a = 1$
• 3 pairs when $a = 2$
• 1 pair when $a = 3$
• 0 pairs when $a = 4,5,6$

Add these:

So, there are 13 ordered pairs $(a,b)$ for which the system has no real solutions.