Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Consider the following system of equations:

$x^2$x2 |
$=$= | $y+14$y+14 |

$y$y |
$=$= | $3x-16$3x−16 |

Loading Graph...

a

By considering the left hand side ($LHS$`L``H``S`) and right hand side ($RHS$`R``H``S`) of each equation, fill in the missing values to verify that the points of intersection on the graphs are solutions of the corresponding system of equations.

First, test the point $\left(1,-13\right)$(1,−13).

$x^2=y+14$ |
||||||
---|---|---|---|---|---|---|

$LHS$LHS |
$=$= | $\left(\editable{}\right)^2$()2 | $RHS$RHS |
$=$= | $\editable{}+14$+14 | |

$=$= | $\editable{}$ | $=$= | $\editable{}$ | |||

$y=3x-16$ |
||||||

$LHS$LHS |
$=$= | $\editable{}$ | $RHS$RHS |
$=$= | $3\times\editable{}-16$3×−16 | |

$=$= | $\editable{}$ |

b

Now test the point $\left(2,-10\right)$(2,−10).

$x^2=y+14$ |
||||||
---|---|---|---|---|---|---|

$LHS$LHS |
$=$= | $\left(\editable{}\right)^2$()2 | $RHS$RHS |
$=$= | $\editable{}+14$+14 | |

$=$= | $\editable{}$ | $=$= | $\editable{}$ | |||

$y=3x-16$ |
||||||

$LHS$LHS |
$=$= | $\editable{}$ | $RHS$RHS |
$=$= | $3\times\editable{}-16$3×−16 | |

$=$= | $\editable{}$ |

Easy

3min

When solving $y=x^2-3x+1$`y`=`x`2−3`x`+1 and $y=x+6$`y`=`x`+6 simultaneously, one point of intersection is at $x=-1$`x`=−1. What is the $y$`y`-coordinate at this point?

Easy

1min

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

Easy

1min

Solve the following equations.

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

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

Easy

1min

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