Question 212·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

0 / 233

In the system of equations

\begin{cases} y = x^2 + ax + 5 \\ y = -x + b \end{cases}

$a$ and $b$ are integers with $0 \le a \le 6$ and $0 \le b \le 10$ . For how many ordered pairs $(a,b)$ does the system have no real solutions?

Your answer

A line and a parabola can cross, touch, or miss each other. If a system asks for no real solutions, set the two expressions for $y$ equal and use the discriminant to test whether they miss. Then count the allowed integer values of the coefficients. A touch gives one solution, so the inequality must be strict.

Hints

Hint 1
At a solution to the system, both equations give the same $y$ -value. Set their right sides equal, then move everything to one side to make a quadratic equation in $x$ .
Hint 2
The discriminant is the part under the square root in the quadratic formula. A quadratic has no real roots only when its discriminant is negative, not when it is zero.
Hint 3
Once you have an upper bound on $b$ for each $a$ , count the nonnegative integers below that bound. If the bound is a whole number, that number itself does not count.

Step-by-step

Use the discriminant, then count integer pairs
Step 1
See what a missed intersection looks like
Try the allowed pair $a=b=0$ . Type $a=0$ , $b=0$ , and the two given equations into Desmos. The graphs do not meet in this window. An intersection is a point on both graphs, so this pair appears to work; algebra will give us a test that doesn't depend on the window.
Step 2
Turn the system into one quadratic
At a point shared by both graphs, the two expressions for $y$ are equal. Set them equal:
$x^2+ax+5=-x+b.$
Subtract the right side to put the equation in quadratic form:
$x^2+(a+1)x+5-b=0.$
Step 3
Test for no real solutions
The discriminant of $x^2+Bx+C=0$ is $B^2-4C$ , the number under the square root in the quadratic formula. Here $B=a+1$ and $C=5-b$ , so no real solutions requires:
$(a+1)^2-4(5-b)<0.$
A negative discriminant means the graphs never meet; zero means they touch once. A touch does not count here.
Step 4
Find the allowed bound on $b$
Expand $-4(5-b)$ :
$(a+1)^2-20+4b<0.$
Move the other terms to the right:
$4b<20-(a+1)^2.$
Divide by $4$ :
$b<\frac{20-(a+1)^2}{4}.$
The strict $<$ sign keeps pairs whose graphs only touch out of the count.
Step 5
Count the possible $b$ values for each $a$
Type $q=[0,1,2,3,4,5,6]$ for the allowed $a$ -values, then $t=\frac{20-(q+1)^2}{4}$ for their bounds on $b$ . Desmos shows $t=[4.75,4,2.75,1,-1.25,-4,-7.25]$ . Since $b$ is a nonnegative integer strictly below its bound, the corresponding counts are $5,4,3,1,0,0,0$ . For example, below $4$ you can use $0,1,2,3$ , but not $4$ . Every positive bound is below $10$ , so the upper limit on $b$ removes nothing else.
Step 6
Add the counts
Type the counts as $c=[5,4,3,1,0,0,0]$ , then type $\operatorname{total}(c)$ . Desmos prints $13$ . Each count belongs to one different value of $a$ , so there are 13 ordered pairs $(a,b)$ for which the system has no real solutions. Grid in 13.

Question 212·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this