Question 78·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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y = x^2 + bx + 3

y = 5x - 3

In the system of equations above, $b$ is a constant. The graphs intersect at two points $(x_1, y_1)$ and $(x_2, y_2)$ , where $x_1 < x_2$ and both $x$ -coordinates are positive. If $x_2 - x_1 = 5$ , what is the value of $b$ ?

A line and a parabola intersect where they give the same $y$ , so setting their expressions equal turns the intersection coordinates into roots of a quadratic. If the problem gives the gap between the roots, use the quadratic’s constant term to get their product, then find the positive roots in Desmos. The middle coefficient gives their sum. Watch the signs: a pair of negative roots can have the right gap too.

Hints

Hint 1
At an intersection, both equations give the same $y$ . Set their right sides equal and move everything to one side. What constant term does the resulting quadratic have?
Hint 2
For a quadratic whose $x^2$ coefficient is $1$ , the constant term is the product of its roots. If the smaller root is $x_1$ , the stated gap tells you how to write the larger root.
Hint 3
Use a regression to find the smaller root from the product, but restrict it to positive values. Then add the roots and compare their sum with the coefficient of $x$ .

Step-by-step

Use the root product and the given gap
Step 1
Make one equation for the intersections
At an intersection, both equations have the same $y$ . So set their right sides equal:
$x^2+bx+3=5x-3.$
Step 2
Put the quadratic in standard form
Subtract $5x-3$ from both sides:
$x^2+bx+3-(5x-3)=0.$
Combine like terms:
$x^2+(b-5)x+6=0.$
Its roots, or solutions for $x$ , are the two intersection $x$ -coordinates.
Step 3
Read the product of the roots
A quadratic with $1$ in front of $x^2$ can be written as $(x-x_1)(x-x_2)$ . Expanding shows why its constant term is the product of its roots:
$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2.$
The constant term above is $6$ , so $x_1x_2=6$ .
Step 4
Find the smaller positive root
The gap of $5$ means $x_2=x_1+5$ . Replace $x_2$ in the product:
$x_1(x_1+5)=6.$
Type $x_1(x_1+5)\sim6\{x_1>0\}$ in Desmos. The $\sim$ tells Desmos to find $x_1$ ; the restriction keeps the root positive, as the problem requires. Under PARAMETERS, Desmos shows $x_1=1$ . Without that restriction, $x_1=-6$ would also fit the product and gap, but both intersection $x$ -coordinates would be negative.
Step 5
Get the sum of the roots
The larger root is $x_1+5$ , so type $x_1+(x_1+5)$ on the next Desmos line. It prints $7$ , the sum $x_1+x_2$ .
Step 6
Use the root sum to find the constant
In $x^2+px+q=0$ , the root sum is $-p$ : the expansion in step 3 has $-(x_1+x_2)$ in front of $x$ . Here $p=b-5$ , so the sum is $5-b$ . Type $5-b\sim7$ in Desmos. Under PARAMETERS, it shows $b=-2$ . That is the constant for two positive intersection $x$ -coordinates five apart. Choice B.

Question 78·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