Question 53·Medium·Nonlinear Equations in One Variable; Systems in Two Variables
Let be a real number. Consider the system of equations
For which value of does the system have exactly one real solution?
When a system involves a line and a parabola and asks for the number of real solutions, think about intersections: usually there are 0, 1, or 2. Set the two expressions for equal to get a quadratic in , then use the discriminant to control the number of real solutions: gives two, gives exactly one, and gives none. On the SAT, go straight to writing the quadratic and discriminant in terms of the parameter, set to the required value (often 0 for “exactly one solution”), and solve for the parameter instead of trying to reason purely from the graph.
Hints
Connect one solution to the graph shapes
You are intersecting a parabola and a line. How many intersection points do they usually have, and what must be special about the line for there to be exactly one intersection point?
Turn the system into a single equation
Set the two expressions for equal to each other to get one equation in terms of and . Then rearrange it into the standard quadratic form .
Use the discriminant condition
For the quadratic you found, write the discriminant (using the coefficients of , , and the constant term). What value must have for there to be exactly one real solution? Set up and solve that equation in terms of .
Desmos Guide
Graph the parabola
In Desmos, enter the parabola equation: y = -x^2 + 4x + 3.
Test each answer choice as a line
For each option , type the corresponding line into Desmos (for example, y = 2x + 0, y = 2x + 2, etc.). After typing each line, look at how many intersection points it has with the parabola (Desmos will show intersection points when you tap on the curves).
Identify the special case
Compare the four lines: find the value of for which the line just touches the parabola at exactly one point (the line is tangent). That value of is the one that makes the system have exactly one real solution.
Step-by-step Explanation
Set the two equations equal
For a solution to the system, both expressions for must be equal:
Move all terms to one side:
Multiply both sides by to make the coefficient positive:
This is a quadratic equation in whose solutions are the -coordinates of the intersection points.
Use the discriminant for the "exactly one real solution" condition
The number of real solutions of a quadratic depends on the discriminant :
- If , there are two distinct real solutions.
- If , there is exactly one real solution (a repeated root).
- If , there are no real solutions.
Here, , , and . So the discriminant is
Simplify the discriminant and set it equal to 0
Simplify :
For the system to have exactly one real solution, we need :
Now solve this equation for . Do not substitute answer choices yet; first find the exact value of that makes this true.
Solve for b and identify the correct choice
Solve
Add to both sides:
Divide both sides by :
So when , the quadratic has exactly one real solution, meaning the line is tangent to the parabola and the system has exactly one real solution. The correct answer choice is C) 4.