Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

We've already learnt how to solve simultaneous 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`). Did you realise that these unknown variables can actually be thought of graphically as the point of intersection between the two equations?

When we have two linear equations (or straight lines), they will only have one point of intersection!

However, when we have a quadratic (parabola) and a linear equation (straight line), there can be:

- two points of intersection

- one point of intersection

- or even no points of intersections (means there are no solutions)

Remember!

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

The examples below help explain how to find these two points of intersection.

Solve the following 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 law to equate these two equations.

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`).

Think: We can use the substitution method to solve these equations simultaneously to find the point of intersection.

Do:

$y$y |
$=$= | $-2x^2+x-12$−2x2+x−12 |
Substitute $x=-5$x=−5 into the equation |

$=$= | $-2\times\left(-5\right)^2+\left(-5\right)-12$−2×(−5)2+(−5)−12 | Now let's simplify | |

$=$= | $-2\times25-5-12$−2×25−5−12 | ||

$=$= | $-50-17$−50−17 | ||

$=$= | $-67$−67 |

So then the point of intersection is $\left(-5,-67\right)$(−5,−67)

Solve the following equations simultaneously.

Equation 1 | $y=x^3$y=x3 |

Equation 2 | $y=-10x^2-25x$y=−10x2−25x |

a) Solve for $x$`x`.

Think: Since $y$`y` is the subject of both equations, we can use the transitive law to equate these two equations.

Do:

$x^3$x3 |
$=$= | $-10x^2-25x$−10x2−25x |

$x^3+10x^2+25x$x3+10x2+25x |
$=$= | $0$0 |

$x\left(x^2+10x+25\right)$x(x2+10x+25) |
$=$= | $0$0 |

$x\left(x+5\right)^2$x(x+5)2 |
$=$= | $0$0 |

$x$x |
$=$= | $0$0,$-5$−5 |

b) Find the value of $y$`y` at the point of intersection where $x=0$`x`=0.

Think: We need to substitute $x=0$`x`=0 into one of the equations.

Do:

$y$y |
$=$= | $x^3$x3 |

$=$= | $0^3$03 | |

$=$= | $0$0 |

Note: We get the same answer when we substitute $x=0$`x`=0 into equation 2.

c) Find the value of $y$`y` at the point of intersection where $x=-5$`x`=−5.

Think: Again I'm going to substitute this value into equation 1.

Do:

$y$y |
$=$= | $x^3$x3 |

$=$= | $\left(-5\right)^3$(−5)3 | |

$=$= | $-125$−125 |

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, $\editable{}$ $)$)