Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

Let's say we have the inequality $x+4<2x-6$`x`+4<2`x`−6.

By now we know that we can use our elementary operations $+$+,$-$−,$\times$×,$\div$÷ to solve inequalities like this in exactly the same way that we solve equations, as long as we remember to reverse the inequality sign whenever we multiply or divide by a negative number. Hence, we could get our answer like this:

$x+4$x+4 |
$<$< | $2x-6$2x−6 |

$x$x |
$<$< | $2x-10$2x−10 |

$-x$−x |
$<$< | $-10$−10 |

$x$x |
$>$> | $10$10 |

But what question are we really being asked when we are asked to solve $x+4<2x-6$`x`+4<2`x`−6? We are being asked for values of $x$`x` at which $x+4$`x`+4 is less than $2x-6$2`x`−6.

We can answer this question graphically by considering two lines $y_1=x+4$`y`1=`x`+4 and $y_2=2x-6$`y`2=2`x`−6.

Solving $x+4<2x-6$`x`+4<2`x`−6 is the same as solving $y_1`y`1<`y`2. In other words, for what values of $x$`x` is the line $y_1$`y`1 *below* the line $y_2$`y`2?

Sure enough, our graph confirms what we already knew: that $x+4<2x-6$`x`+4<2`x`−6 for $x>10$`x`>10.

Now, what if we are given an inequality involving a quadratic, such as $x^2+2x-8>0$`x`2+2`x`−8>0? Our elementary operations alone will be insufficient to solve this algebraically. We will need to use the above method to find our solutions using a graph.

Let's use what we already know about quadratics to graph $y=x^2+2x-8$`y`=`x`2+2`x`−8. We know that the parabola will be concave up, since the coefficient of $x^2$`x`2 is positive. We also get a $y$`y`-intercept of $y=-8$`y`=−8 after substituting $x=0$`x`=0. We can use our vertex formula $x=\frac{-b}{2a}$`x`=−`b`2`a` to find the axis of symmetry where our turning point lies, which we know will be a minimum turning point due to the concavity.

$x$x |
$=$= | $\frac{-b}{2a}$−b2a |

$=$= | $\frac{-2}{2\times1}$−22×1 | |

$=$= | $-1$−1 |

Last of all, we can also factorise the quadratic to find our $x$`x`-intercepts.

$x^2+2x-8$x2+2x−8 |
$=$= | $0$0 |

$\left(x+4\right)\left(x-2\right)$(x+4)(x−2) |
$=$= | $0$0 |

$x$x |
$=$= | $-4$−4,$2$2 |

Now we can graph our parabola.

Being asked to find solutions to $x^2+2x-8>0$`x`2+2`x`−8>0 is exactly the same as being asked to find values of $x$`x` at which the graph of $y=x^2+2x-8$`y`=`x`2+2`x`−8 is above the line $y=0$`y`=0 (the $x$`x`-axis). In other words, when is $y=x^2+2x-8$`y`=`x`2+2`x`−8 positive?

We can see from the graph here that unlike our earlier inequality involving lines only, our answer has to be split into two parts, $x<-4$`x`<−4 and $x>2$`x`>2. No wonder we couldn't solve this algebraically using elementary operations.

What if we flipped the inequality sign in the question? So, solve $x^2+2x-8<0$`x`2+2`x`−8<0. Now we are being asked when the graph $y=x^2+2x-8$`y`=`x`2+2`x`−8 is *negative*. Now our solutions will be $x>-4$`x`>−4 and $x<2$`x`<2. This can be written as $-4`x`$<$<$2$2.

You may have noticed by now that if a parabola has two distinct roots (when its discriminant is positive), it will always be on one side of the $x$`x`-axis between the two intercepts, and on the *other *side of the $x$`x`-axis outside the two intercepts. Parabolas are either positive between the two intercepts and negative outside of them, or negative between the two intercepts and positive outside of them.

In other words, for any parabola with two distinct $x$`x`-intercepts $a$`a` and $b$`b`, we only ever have one of two possible situations. We'll assume here that $a`a`<`b`.

$x$x |
$xx<a |
$aa<x$<$<$b$b |
$x>b$x>b |

$y$y |
$+$+ | $-$− | $+$+ |

OR

$x$x |
$xx<a |
$aa<x$<$<$b$b |
$x>b$x>b |

$y$y |
$-$− | $+$+ | $-$− |

So, if we have our two $x$`x`-intercepts, we only need a test value for each interval to determine if $y$`y` is positive or negative in that interval.

Let's use this method to solve our previous quadratic inequality $x^2+2x-8>0$`x`2+2`x`−8>0. We know our $x$`x`-intercepts, so let's choose some test values for each interval to find when $y$`y` is positive.

$x$x |
$x<-4$x<−4 |
$-4x$<$<$2$2 |
$x>2$x>2 |

Test Value |
$x=-5$x=−5 |
$x=0$x=0 |
$x=3$x=3 |

$y$y |

$y$y$($($-5$−5$)$) |
$=$= | $\left(-5\right)^2+2\times\left(-5\right)-8$(−5)2+2×(−5)−8 |

$=$= | $7$7 | |

$y$y$($($0$0$)$) |
$=$= | $0^2+2\times0-8$02+2×0−8 |

$=$= | $-8$−8 | |

$y$y$($($3$3$)$) |
$=$= | $3^2+2\times3-8$32+2×3−8 |

$=$= | $7$7 |

Hence, we can fill in the signs in our table as follows.

$x$x |
$x<-4$x<−4 |
$-4x$<$<$2$2 |
$x>2$x>2 |

Test Value |
$x=-5$x=−5 |
$x=0$x=0 |
$x=3$x=3 |

$y$y |
$+$+ | $-$− | $+$+ |

Therefore, the solution to our inequality $x^2+2x-8>0$`x`2+2`x`−8>0 is $x<-4$`x`<−4 or $x>2$`x`>2 as expected.

Suppose we want to solve $\left(x+2\right)^2>2x+12$(`x`+2)2>2`x`+12. One approach would be to go through the laborious task of graphing the parabola on the left and the line on the right together to examine when the parabola lies above the line. However, there is no need for this.

Actually, we can rearrange this inequality as follows:

$\left(x+2\right)^2$(x+2)2 |
$>$> | $2x+12$2x+12 |

$x^2+4x+4$x2+4x+4 |
$>$> | $2x+12$2x+12 |

$x^2+2x-8$x2+2x−8 |
$>$> | $0$0 |

As it turns out, the inequality $\left(x+2\right)^2>2x+12$(`x`+2)2>2`x`+12 is equivalent to the inequality $x^2+2x-8>0$`x`2+2`x`−8>0 from before, and will have the same solutions!

Whenever we see a quadratic inequality, it is a good idea (and much easier!) to rearrange it in this way to have a zero on the right hand side.

Consider the graph of $y=f\left(x\right)$`y`=`f`(`x`).

Loading Graph...

Find the values of $x$

`x`for which $f\left(x\right)=0$`f`(`x`)=0. Write both solutions on the same line separated by a comma.For what values of $x$

`x`is $f\left(x\right)<0$`f`(`x`)<0?Give your answer as an inequality.

For what values of $x$

`x`is $f\left(x\right)>0$`f`(`x`)>0?Give your answer as an inequality.

What is the $x$

`x`-coordinate of the vertex of $f\left(x\right)$`f`(`x`)?

Consider the inequality $x^2-3x-10\ge0$`x`2−3`x`−10≥0.

First factorise $x^2-3x-10$

`x`2−3`x`−10.For what values of $x$

`x`is $x^2-3x-10=0$`x`2−3`x`−10=0? Leave your solutions on the same line separated by a comma.To solve the inequality, Micky splits up the entire number line into three intervals, and tests a value from each interval to see if it satisfies the inequality $x^2-3x-10\ge0$

`x`2−3`x`−10≥0.Interval A ($x<-2$ `x`<−2)$x=-2$ `x`=−2Interval B ($-2

−2< `x`$<$<$5$5)$x=5$ `x`=5Interval C ($x>5$ `x`>5)Value to test Value to test Value to test $x=-3$ `x`=−3$x=1$ `x`=1$x=7$ `x`=7Which values of $x$

`x`satisfy the inequality $x^2-3x-10\ge0$`x`2−3`x`−10≥0? Select all that apply.$x=7$

`x`=7A$x=-3$

`x`=−3B$x=1$

`x`=1CHence, state the solution to the inequality $x^2-3x-10\ge0$

`x`2−3`x`−10≥0. You may use the keywords "and"/"or".

Consider the inequality $x^2-5x\ge3x-6$`x`2−5`x`≥3`x`−6.

First, solve the equation $x^2-5x=3x-6$

`x`2−5`x`=3`x`−6.Hence, solve the inequality $x^2-5x\ge3x-6$

`x`2−5`x`≥3`x`−6. Leave your answer in exact form. You may use the keywords "and"/"or".