Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

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

This is a statement about the variable $x$`x`. We're not asked to solve this equation, and we don't need to do any balancing of sides. We're simply stating that it is true. In fact, it is always true no matter what the value of $x$`x` is. The left hand side and the right hand are equivalent expressions. (Expand and simplify the right hand side to see that this is the case!)

Equations that include variables and are always true are called identities. The above statement is an example of a quadratic identity.

It can be useful to rewrite a quadratic expression as a string of terms with binomials, as above. This will come in handy further down the track when we're dealing with rational functions with quadratic expressions in the numerator.

Let's suppose we have the quadratic expression $4x^2+3x+7$4`x`2+3`x`+7. Every quadratic expression has an endless number of equivalent expressions, so the amount of identities we could come up with for this quadratic expression are endless. However, what if we are *only *interested in finding some equivalent expression of the form $a\left(x+4\right)^2+b\left(x-2\right)+c$`a`(`x`+4)2+`b`(`x`−2)+`c`? In other words, what if we're *forced *to find a particular kind of identity?

Our job now is to somehow find $a$`a`, $b$`b` and $c$`c` for the identity $a\left(x+4\right)^2+b\left(x-2\right)+c=4x^2+3x+7$`a`(`x`+4)2+`b`(`x`−2)+`c`=4`x`2+3`x`+7. You may ask, "But I thought identities weren't about finding solutions?" It's a good question.

$a$`a`, $b$`b` and $c$`c` now represent a whole range of different expressions on the left hand side of the equation. Only one of those left hand sides will make an identity. The rest will make regular equations that will only be true for some solutions of $x$`x`. We want to find the identity. This means we have to solve for $a$`a`, $b$`b` and $c$`c`, momentarily treating $x$`x` as if it were a constant.

There are two main ways we can go about this: by equating the coefficients, and by substitution.

Let's get the left hand side expression $a\left(x+4\right)^2+b\left(x-2\right)+c$`a`(`x`+4)2+`b`(`x`−2)+`c` and expand it out.

$a\left(x+4\right)^2+b\left(x-2\right)+c$a(x+4)2+b(x−2)+c |
$=$= | $a\left(x^2+8x+16\right)+b\left(x-2\right)+c$a(x2+8x+16)+b(x−2)+c |

$=$= | $ax^2+8ax+16a+bx-2b+c$ax2+8ax+16a+bx−2b+c |

Now, let's rearrange this string of terms so that the coefficients of $x$`x` are grouped together.

$a\left(x+4\right)^2+b\left(x-2\right)+c$a(x+4)2+b(x−2)+c |
$=$= | $ax^2+16a+\left(8a+b\right)x-2b+c$ax2+16a+(8a+b)x−2b+c |

And now the same for the constant term.

$a\left(x+4\right)^2+b\left(x-2\right)+c$a(x+4)2+b(x−2)+c |
$=$= | $ax^2+\left(8a+b\right)x+\left(16a-2b+c\right)$ax2+(8a+b)x+(16a−2b+c) |

For $a\left(x+4\right)^2+b\left(x-2\right)+c=4x^2+3x+7$`a`(`x`+4)2+`b`(`x`−2)+`c`=4`x`2+3`x`+7 to be an identity, we want $a\left(x+4\right)^2+b\left(x-2\right)+c$`a`(`x`+4)2+`b`(`x`−2)+`c` to be an equivalent expression to $4x^2+3x+7$4`x`2+3`x`+7. This means, given our expansion above, that we want $\left(a\right)x^2+\left(8a+b\right)x+\left(16a-2b+c\right)$(`a`)`x`2+(8`a`+`b`)`x`+(16`a`−2`b`+`c`) to be an equivalent expression to $4x^2+3x+7$4`x`2+3`x`+7.

In order for this to be the case, the coefficients of the $x^2$`x`2, $x$`x`, and constant terms between the two expressions *must *be equal. In other words,

$a$a |
$=$= | $4$4 |

$8a+b$8a+b |
$=$= | $3$3 |

$16a-2b+c$16a−2b+c |
$=$= | $7$7 |

So we now have a set of simultaneous equations in $a$`a`, $b$`b` and $c$`c` that we can solve. $a$`a` has been given to us, so we can just use our substitution method into the second equation to get

$8\times4+b$8×4+b |
$=$= | $3$3 |

$32+b$32+b |
$=$= | $3$3 |

$b$b |
$=$= | $3-32$3−32 |

$b$b |
$=$= | $-29$−29 |

and then substitute this into our third equation to get

$16\times4-2\times\left(-29\right)+c$16×4−2×(−29)+c |
$=$= | $7$7 |

$64+58+c$64+58+c |
$=$= | $7$7 |

$122+c$122+c |
$=$= | $7$7 |

$c$c |
$=$= | $-115$−115 |

And so we have our identity $4\left(x+4\right)^2-29\left(x-2\right)-115=4x^2+3x+7$4(`x`+4)2−29(`x`−2)−115=4`x`2+3`x`+7.

Another way we could have arrived at this same conclusion is by substituting in clever values of $x$`x` (after all, any value should be giving a true equation) to find $a$`a`, $b$`b` or $c$`c`.

$a=4$`a`=4 was easy to find by equating coefficients. If we substitute $x=2$`x`=2 into $4\left(x+4\right)^2+b\left(x-2\right)+c=4x^2+3x+7$4(`x`+4)2+`b`(`x`−2)+`c`=4`x`2+3`x`+7, the $b\left(x-2\right)$`b`(`x`−2) becomes zero and we have

$4\left(2+4\right)^2+c$4(2+4)2+c |
$=$= | $4\times2^2+3\times2+7$4×22+3×2+7 |

$4\times36+c$4×36+c |
$=$= | $4\times4+6+7$4×4+6+7 |

$144+c$144+c |
$=$= | $29$29 |

$c$c |
$=$= | $29-144$29−144 |

$c$c |
$=$= | $-115$−115 |

Now knowing $c$`c`, we could substitute $x=-4$`x`=−4 to make the $4\left(x+4\right)^2$4(`x`+4)2 term zero and find $b$`b`.

$b\left(-4-2\right)-115$b(−4−2)−115 |
$=$= | $4\left(-4\right)^2+3\times\left(-4\right)+7$4(−4)2+3×(−4)+7 |

$-6b-115$−6b−115 |
$=$= | $4\times16-12+7$4×16−12+7 |

$-6b-115$−6b−115 |
$=$= | $59$59 |

$-6b$−6b |
$=$= | $174$174 |

$b$b |
$=$= | $-29$−29 |

Both methods will lead to the correct identity.

Notice that when finding the quadratic identity $a\left(x+4\right)^2+b\left(x-2\right)+c=4x^2+3x+7$`a`(`x`+4)2+`b`(`x`−2)+`c`=4`x`2+3`x`+7, there was no balancing involved of this equation. Remember this when going through problems. Trying to balance this equation will only lead to confusion!

Consider the equation $ax\left(x+6\right)+b\left(x+7\right)+c=3x^2+13x+7$`a``x`(`x`+6)+`b`(`x`+7)+`c`=3`x`2+13`x`+7.

Rearrange the left hand side of the above equation such that the coefficients of $x$

`x`are grouped together and the constant terms are grouped together.Equate the coefficients of $x^2$

`x`2 to find $a$`a`. Write each line of working as an equation.Now equate the coefficients of $x$

`x`to find the value of $b$`b`. Write each line of working as an equation.Similarly, equate the constant term on the left hand side and the constant term on the right hand side, and use your answers to the previous part to solve for $c$

`c`.

Consider the equation $a\left(x-7\right)^2+b\left(x-3\right)+c=2x^2+4x+6$`a`(`x`−7)2+`b`(`x`−3)+`c`=2`x`2+4`x`+6.

Equate the coefficients of $x^2$

`x`2 to find the value of $a$`a`. Write each line of working as an equation.By substituting $x=3$

`x`=3 into the equation, solve for the value of $c$`c`. Write each line of working as an equation.Now find the value of $b$

`b`by substituting $x=7$`x`=7 into the equation. Write each line of working as an equation.

Consider the identity $a\left(x-4\right)+b\left(7-x\right)=3x-4$`a`(`x`−4)+`b`(7−`x`)=3`x`−4.

Rearrange the left hand side of the above equation such that the coefficients of $x$

`x`are grouped together and the constant terms are grouped together.Now form an equation involving $a$

`a`and $b$`b`by equating the coefficient of $x$`x`on the left hand side and the coefficient of $x$`x`on the right hand side.Similarly, form an equation involving $a$

`a`and $b$`b`by equating the constant term on the left hand side and the constant term on the right hand side.We have the following equations:

Equation 1 $a-b=3$ `a`−`b`=3Equation 2 $4a-7b=4$4 `a`−7`b`=4First solve for $b$

`b`.Now solve for $a$

`a`.