2. Complex Numbers & Quadratics

Lesson

In Algebra 1, we learned how to solve systems of linear equations using the elimination method and the substitution method. When we solved equations simultaneously, we found two unknown variables (which were usually called $x$`x` and $y$`y`). Remember that the solution to a system of equations can also be thought of graphically as the point of intersection between the two equations.

When we have two linear equations (or straight lines), they can only have one, zero, or infinitely many points of intersection. That is, two lines will never intersect at only two (or three, or four) separate points. However, when the system of equations includes a quadratic (parabola) and a linear equation (straight line), there can exist:

- two points of intersection

- one point of intersection

- or even no points of intersection (this means the system has no solution)

Remember!

Solving a linear and a quadratic equation simultaneously may produce 0, 1 or 2 solutions.

Use the examples below to explore how to find the points of intersection between one quadratic equation and one linear equation, and solve this type of system of equations.

Solve the following system of equations

Equation 1 | $y=x^2$y=x2 |

Equation 2 | $y=36$y=36 |

a) Solve for the value of $x$`x` that satisfies both equations.

Think: Since y is the subject of both equations, we can use the transitive property to set the equations equal to each other.

Do:

$x^2$x2 |
$=$= | $36$36 |

$x$x |
$=$= | $\pm\sqrt{36}$±√36 |

$x$x |
$=$= | $\pm6$±6 |

b) Hence find the points of intersection of the two graphs. Write the coordinates in the form $\left(a,b\right)$(`a`,`b`).

Think: We found two $x$`x` values in part A, so we will need to substitute both these values into the equation to get both possible $y$`y` values. In other words, we should have two points of intersection.

Do:

When $x=6$`x`=6, $y=6^2$`y`=62$=$=$36$36

When $x=-6$`x`=−6, $y=\left(-6\right)^2$`y`=(−6)2$=$=$36$36

Or, if we just looked at equation 2, we'd see that $y=36$`y`=36.

So the two points of intersection are $\left(6,36\right)$(6,36) and $\left(-6,36\right)$(−6,36).

Where does the vertical line $x=-5$`x`=−5 intersect the curve $y=-2x^2+x-12$`y`=−2`x`2+`x`−12?

Give the point of intersection in the form $\left(a,b\right)$(

`a`,`b`).

Use the following two problems to explore the methods for solving a system of two equations where both equations are quadratic.

If $f(x)=x^2+1$`f`(`x`)=`x`2+1 and $g(x)=-x^2+x+4$`g`(`x`)=−`x`2+`x`+4, we want to find the value or values of $x$`x` such that $f(x)=g(x)$`f`(`x`)=`g`(`x`). We can begin by assuming the two functions have a shared value, which means we can write

$x^2+1=-x^2+x+4$`x`2+1=−`x`2+`x`+4.

After collecting like terms, we have the quadratic equation $2x^2-x-3=0$2`x`2−`x`−3=0 and with the help of the quadratic formula or factoring to arrive at $(2x-3)(x+1)=0$(2`x`−3)(`x`+1)=0, we see that the solutions are

$x=\frac{3}{2}$`x`=32 and $x=-1$`x`=−1

Then, we substitute these $x$`x`-values into either one of the original functions to find the corresponding function values. Thus,

$f\left(\frac{3}{2}\right)=\left(\frac{3}{2}\right)^2+1=\frac{13}{4}$`f`(32)=(32)2+1=134 and

$f\left(-1\right)=\left(-1\right)^2+1=2$`f`(−1)=(−1)2+1=2

We conclude that the points of intersection of the graphs of the two functions are

$(-1,2)$(−1,2) and

$\left(\frac{3}{2},\frac{13}{4}\right)$(32,134)

The graphs are shown below.

A system of quadratic functions may be given in the form of a pair of equations of the form $y=ax^2+bx+c$`y`=`a``x`2+`b``x`+`c`. It is possible for a system of two quadratic functions to have no points in common, one common point, or two common points.

Two quadratic functions are given by

$y$y |
$=$= | $x^2+x-1$x2+x−1 |

$y$y |
$=$= | $4x^2+3x-\frac{2}{3}$4x2+3x−23 |

Their graphs are shown below.

The graphs appear to intersect at one point. We check this algebraically.

We can eliminate $y$`y` from the two equations by subtracting the first equation from the second. This gives the quadratic equation $3x^2+2x+\frac{1}{3}=0$3`x`2+2`x`+13=0. According to the quadratic formula, this has the single solution $x=-\frac{1}{3}$`x`=−13. At this value of $x$`x`, the corresponding $y$`y`-value in the original equations is $-\frac{11}{9}$−119.

Solve the following equations.

Equation 1 | $y=x^2$y=x2 |

Equation 2 | $y=9$y=9 |

Solve for the values of $x$

`x`that satisfy both equations.Hence find the points of intersection of the two graphs. Write the coordinates in the form $\left(a,b\right)$(

`a`,`b`).$($( $-3$−3, $\editable{}$ $)$)

$($( $3$3, $\editable{}$ $)$)

In business, supply and demand functions are ways of expressing the supply or demand of a commodity as a function of its unit price. Market equilibrium is the term that describes when the supply is equal to the demand. Consider the supply and demand functions for a new power drill below, where $y$`y` is in dollars and $x$`x` is in thousands of units supplied.

Supply function | $y=4x^2$y=4x2 |

Demand function | $y=-6x+5x^2$y=−6x+5x2 |

We want to find the equilibrium quantity and the corresponding price by solving the system of equations. Start by solving for $x$

`x`.Find the value of $y$

`y`when $x=0$`x`=0.Find the value of $y$

`y`when $x=6$`x`=6.Hence, the equilibrium quantity occurs when $\editable{}$ thousand units are being produced, which has a corresponding price of $\editable{}$ dollars per unit.

Consider the system of equations

$y=x^2+4x+1$`y`=`x`2+4`x`+1

$y=-x^2+7$`y`=−`x`2+7

Shade the region contained inside the graphs (including the boundaries).

Loading Graph...What are the coordinates of the points of intersection?

Write the coordinates of both points on the same line separated by a comma.

Using graphs, tables, or successive approximations, show that the solution to the equation f(x) = g(x) is the x-value where the y-values of f(x) and g(x) are the same.