Quadratic Equations

New Zealand

Level 7 - NCEA Level 2

Lesson

We know by now that if we're given a quadratic equation $x^2+3x+2=0$`x`2+3`x`+2=0, we can factorise this as $\left(x+1\right)\left(x+2\right)=0$(`x`+1)(`x`+2)=0 to find solutions $x=-1$`x`=−1,$-2$−2.

In most of the work we've been doing with equations, we've been asked to solve them, and problems have involved the following process:

**1)** We have been given some **equation**.

**2)** We have been asked to find **solutions** that make the equation true.

However, we can also encounter problems where we have to find an equation. These problems will involve the reverse of the above process:

**1)** We have been given **solutions**.

**2)** We have been asked to find some **equation** that is true for these solutions.

We're going to practice finding quadratic equations from given solutions.

Let's say we have the solutions $x=\pm4$`x`=±4 and are asked to find some quadratic equation that is true for these solutions. Notice that $x^2=16$`x`2=16 is true for both solutions. This means, after rearranging to general form, that $x^2-16=0$`x`2−16=0 is a quadratic equation for these solutions.

Is this the *only* equation for these solutions? $-x^2=-16$−`x`2=−16 is also true for both solutions, which gives a general form quadratic equation of $-x^2+16=0$−`x`2+16=0. This comes from multiplying both sides of $x^2-16=0$`x`2−16=0 by $-1$−1. In fact, we could multiply both sides of $x^2-16=0$`x`2−16=0 by any number we like to get equivalent equations $2x^2-32=0$2`x`2−32=0, $3x^2-32=0$3`x`2−32=0, etc.

This leads to an important observation. For any given quadratic equation, there are at most two solutions, but for any two given solutions, there are an infinite number of quadratic equations.

In practice, there will be other conditions specific to the situation that will allow us to determine the unique quadratic equation that we want. These conditions might be that the quadratic also pass through another point, like the vertex or $y$`y`-intercept.

Recall that every monic quadratic equation (equations of the form $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0 where $a=1$`a`=1) with two solutions $\alpha$`α` and $\beta$`β` can be factorised to $\left(x-\alpha\right)\left(x-\beta\right)=0$(`x`−`α`)(`x`−`β`)=0. This means that for any pair of solutions $\alpha$`α` and $\beta$`β`, there is only one unique monic quadratic $\left(x-\alpha\right)\left(x-\beta\right)=0$(`x`−`α`)(`x`−`β`)=0.

Suppose we're asked to find the monic quadratic equation for the solutions $x=2$`x`=2 and $x=-5$`x`=−5. We know that the equation must be of the form $\left(x-2\right)\left(x-\left(-5\right)\right)=0$(`x`−2)(`x`−(−5))=0 or $\left(x-2\right)\left(x+5\right)=0$(`x`−2)(`x`+5)=0. We can represent this in expanded form as $x^2+3x-10=0$`x`2+3`x`−10=0.

Form a monic quadratic equation which has solutions $x=-3$`x`=−3 and $x=-5$`x`=−5. Express the equation in expanded form.

Again, if we are given a particular vertex, there are an infinite number of parabolic functions that will have that particular vertex. However, there will only be one *monic *parabolic function.

Recall that every monic parabolic function with vertex $h,k$`h`,`k` can be written as $y=\left(x-h\right)^2+k$`y`=(`x`−`h`)2+`k`, the unique monic quadratic function for this particular vertex.

So, if we are asked to find the monic parabolic function with vertex $3,4$3,4, we know it will be $y=\left(x-3\right)^2+4$`y`=(`x`−3)2+4, which can then be expanded to general form as $y=x^2-6x-5$`y`=`x`2−6`x`−5.

A parabola of the form $y=\left(x-h\right)^2+k$`y`=(`x`−`h`)2+`k` is symmetrical about the line $x=2$`x`=2, and its vertex lies $6$6 units below the $x$`x`-axis.

Determine the equation of the parabola.

Graph the parabola.

Loading Graph...

We found above that if we're given a pair of solutions or a vertex, then there is a unique monic equation to be found, but an infinite number of non-monic equations. However, there is a way around this.

As long as we're given an additional point that lies somewhere on the parabola or satisfies the equation, there will only be one unique non-monic equation to find.

Recall that every quadratic expression can be written in general factorised form as $a\left(x-\alpha\right)\left(x-\beta\right)$`a`(`x`−`α`)(`x`−`β`) for the $x$`x`-intercepts or $a\left(x-h\right)^2+k$`a`(`x`−`h`)2+`k` for the turning point, where $a$`a` is simply the coefficient from the general form $ax^2+bx+c=0$`a``x`2+`b``x`+`c`=0.

We can use the solutions or vertex to put the equation in this form, and then solve for $a$`a` by substituting in our additional point.

Determine the equation of a parabola whose $x$`x`-intercepts are $-10$−10 and $4$4, and whose $y$`y`-intercept is $-40$−40.

Express the equation in the following form, for some value of $a$

`a`.$y=a\left(x+\editable{}\right)\left(x-\editable{}\right)$

`y`=`a`(`x`+)(`x`−)Note: you do not need to find the value of $a$

`a`at this point.Hence determine the value of $a$

`a`.Hence state the equation of the parabola in the form $y=x^2+\ldots$

`y`=`x`2+…

Occasionally, we might be given an incomplete equation such as $y=5x^2+bx+9$`y`=5`x`2+`b``x`+9 and be asked to find the full equation by figuring out $b$`b`. To do this, the question will give us some extra condition such as the number of $x$`x`-intercepts (which would require us to use the discriminant), or the axis of symmetry (which would require us to use the vertex formula).

The parabola $y=2x^2+bx+1$`y`=2`x`2+`b``x`+1 has its axis of symmetry at $x=-1$`x`=−1. Find the value of $b$`b`.