Quadratic Equations

New Zealand

Level 7 - NCEA Level 2

Lesson

Suppose we are asked the solve the equation $x^4-2x^2+1=0$`x`4−2`x`2+1=0. At first glance, given the presence of the $x^4$`x`4 term, it might seem as though we don't yet know how to solve this kind of equation.

However, if we remember our index laws and notice that $x^4=\left(x^2\right)^2$`x`4=(`x`2)2, we can rewrite the equation as $\left(x^2\right)^2-2x^2+1=0$(`x`2)2−2`x`2+1=0. Does this look familiar at all?

What if we let $m=x^2$`m`=`x`2 and substitute this into the equation? There is nothing wrong with doing this. We are changing names, but the equation is still the same. The equation is now written as $m^2-2m+1=0$`m`2−2`m`+1=0. Success! The equation is in fact just a quadratic in disguise. We have reduced the equation to a quadratic.

But what happened to the $x^2$`x`2? We started out wanting to solve an equation for $x$`x`, and now we have to solve an equation for $m$`m`.

Well, $m$`m` is simply standing in for $x^2$`x`2 in the meantime to make it easier for us to do our working. Eventually, when we don't need $m$`m` anymore, we will change back to $x^2$`x`2.

Now, let's solve the equation.

$m^2-2m+1$m2−2m+1 |
$=$= | $0$0 |
Let's factorise this. |

$(m-1)^2$(m−1)2 |
$=$= | $0$0 |
Which we can solve for $m$ |

$m$m |
$=$= | $1$1 |
Substitute back in for $m=x^2$ |

$x^2$x2 |
$=$= | $1$1 |
And then we can solve for our final solutions for $x$ |

$x$x |
$=$= | $\pm1$±1 |

That's right, the roots that we found for $m$`m` are equations in themselves! We then solve these to get our overall solutions $x=1$`x`=1 and $x=-1$`x`=−1 to our equation $x^4-2x^2+1=0$`x`4−2`x`2+1=0. Check that these values of $x$`x` satisfy the equation using a calculator.

What if we are asked to solve the equation $2^{2x}-12\times2^x+32=0$22`x`−12×2`x`+32=0? It looks like a hopeless mess of terms, expressions and operations that could take forever to unravel in order to find the value of $x$`x` which satisfies the equation.

Again, remember the index laws and notice that $2^{2x}=\left(2^x\right)^2$22`x`=(2`x`)2. We can therefore rewrite the equation as $\left(2^x\right)^2-12\left(2^x\right)+32=0$(2`x`)2−12(2`x`)+32=0.

Now we can use a substitution $m=2^x$`m`=2`x`.

We proceed as before.

$m^2-12m+32$m2−12m+32 |
$=$= | $0$0 |

$\left(m-4\right)\left(m-8\right)$(m−4)(m−8) |
$=$= | $0$0 |

$m$m |
$=$= | $4,8$4,8 |

$2^x$2x |
$=$= | $4,8$4,8 |

$x$x |
$=$= | $2,3$2,3 |

Again, check that these values of $x$`x` satisfy the equation using a calculator.

Eventually, once reducible quadratics become easier, we may not need to use a substitution. In the above equation, we could have very well gone straight from $2^{2x}-12\times2^x+32=0$22`x`−12×2`x`+32=0 to $\left(2^x-4\right)\left(2^x-8\right)=0$(2`x`−4)(2`x`−8)=0. This will come with practice.

Solve the following equation for $x$`x` by substituting in $m=4^x$`m`=4`x`.

$4^{2x}-65\times4^x+64=0$42`x`−65×4`x`+64=0

Use a comma to separate multiple solutions.

If instead the equation to solve was $\log_2^2x-5\log_2x+6=0$`l``o``g`22`x`−5`l``o``g`2`x`+6=0, we would use a substitution $m=\log_2x$`m`=`l``o``g`2`x` to factorise the equation.

$m^2-5m+6$m2−5m+6 |
$=$= | $0$0 |
Let's factorise this. |

$\left(m-2\right)\left(m-3\right)$(m−2)(m−3) |
$=$= | $0$0 |
Which we can solve for $m$ |

$m$m |
$=$= | $2,3$2,3 |
Substitute back in for $m=\log_2x$ |

$\log_2x$log2x |
$=$= | $2,3$2,3 |
And then we can solve for our final solutions for $x$ Remember that $\log_2x=3$ |

$x$x |
$=$= | $4,8$4,8 |

Solve $3\log^2x-5\log x=12$3`l``o``g`2`x`−5`l``o``g``x`=12.

1. Just as quadratics don't always have solutions, the equations we get from the roots of our quadratic might not always have solutions. For instance, if we get roots $m=-2$`m`=−2 and $m=-4$`m`=−4, but $m=2^x$`m`=2`x`, then $x$`x` can't have any solutions because $2^x$2`x` must always be positive!

2. Sometimes, particularly when $e^x$`e``x` and $\ln x$`l``n``x` are involved, we will have to leave our final answers in exact form.

3. The reducible quadratic might not be obvious at first. We may have to use the index and log laws to rewrite the equation at the start until we get an equation that can be reduced.

Solve $e^{2x}-8e^x+7=0$`e`2`x`−8`e``x`+7=0, expressing your solutions in exact form.

Find the value of $x$`x` for which $\left(\log_3x\right)^2-\log_3\left(x^4\right)+4=0$(`l``o``g`3`x`)2−`l``o``g`3(`x`4)+4=0.