Question 123·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

0 / 233

The system of equations is given by

$y = 2x^2 - 5x + 4$

$3x + 2y = 10$

If $(x_1, y_1)$ and $(x_2, y_2)$ are the two solutions to the system of equations above, what is the value of $y_1 y_2$ ?

A squared variable paired with a line signals a nonlinear system: the graphs may meet at two points. Graph both equations in Desmos and click both intersections to get the two solutions. Then use the coordinates the question asks for, rather than submitting an intersection or rounding its coordinates too soon.

Hints

Hint 1
An intersection is a point on both graphs, so it satisfies both equations. Enter each equation in Desmos exactly as written, then look for both places where the line meets the curve.
Hint 2
In a point written $(x,y)$ , the second number is $y$ . After clicking both intersections, which two numbers should you use for $y_1y_2$ ?
Hint 3
Multiply the two $y$ -coordinates before rounding. If Desmos shows a decimal product and the choices are fractions, use the fraction button beside its result.

Step-by-step

Approach 1: Graph both equations and multiply the y-coordinates
Step 1
Find both solutions on the graph
Type $y=2x^2-5x+4$ and $3x+2y=10$ on separate lines in Desmos. Click both intersections. Desmos shows $(-0.25,5.375)$ and $(2,2)$ . Each intersection satisfies both equations, and its second coordinate is the $y$ -value you need. So $y_1$ and $y_2$ are $5.375$ and $2$ , in either order.
Step 2
Multiply the requested coordinates
Type $5.375\cdot2$ . Desmos shows $10.75$ ; its fraction button shows $\frac{43}{4}$ . Don't multiply the $x$ -coordinates or round first: the question asks for the product of the two $y$ -coordinates. So $y_1y_2=\frac{43}{4}$ . Choice C.

Approach 2: Build a quadratic whose roots are the y-coordinates
Step 1
Write x in terms of y
To get an equation whose solutions are the two $y$ -coordinates, solve the line for $x$ . Subtract $2y$ :
$3x=10-2y.$
Divide by $3$ :
$x=\frac{10-2y}{3}.$
Step 2
Substitute into the parabola
At either solution, the same $x$ must satisfy the parabola. Replace $x$ with $\frac{10-2y}{3}$ to leave only $y$ :
$y=2\left(\frac{10-2y}{3}\right)^2-5\left(\frac{10-2y}{3}\right)+4.$
Step 3
Put the equation in quadratic form
Multiply by $9$ to clear the denominators:
$9y=2(10-2y)^2-15(10-2y)+36.$
Expand:
$9y=200-80y+8y^2-150+30y+36.$
Combine like terms:
$9y=8y^2-50y+86.$
Subtract $9y$ so the quadratic equals zero:
$0=8y^2-59y+86.$
Step 4
Read the product from the coefficients
A root is a value that makes the quadratic equal $0$ , so its roots are $y_1$ and $y_2$ . In $ay^2+by+c=0$ , the roots multiply to $c/a$ : expanding $a(y-y_1)(y-y_2)$ makes its constant term $ay_1y_2=c$ . Here,
$y_1y_2=\frac{86}{8}=\frac{43}{4}.$
So the two solutions' $y$ -coordinates have product $\frac{43}{4}$ . Choice C.

Question 123·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