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# Roots to Determine a Quadratic Function

## Compose quadratic functions from given roots

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Practice Roots to Determine a Quadratic Function
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Roots to Determine a Quadratic Function

What quadratic function has roots of $2$ and $7$ ? Does more than one function have these roots?

### Guidance

If $x=2$ and $x=5$ are roots of a quadratic function, then $(x - 2)$ and $(x-5)$ must have been factors of the quadratic equation. Therefore, a quadratic function with roots $2$ and $5$ is:

$y&=(x-2)(x-5)\\y&=x^2-7x+10$

Note that there are many other functions with roots of $2$ and $5$ . These functions will all be multiples of the function above. The function below would also work:

$y&=2(x-2)(x-5)\\y&=2x^2-14x+20$

If you don't know the solutions of a quadratic equation, but you know the sum and the product of the solutions, you can find the equation. If $r_1$ and $r_2$ are the solutions to the quadratic equation $ax^2+bx+c=0$ , then

${\color{red}r_1 + r_2 = -\frac{b}{a} \ and \ r_1 \times r_2 = \frac{c}{a}}$ .

The quadratic can be rewritten as:

${\color{red}x^2 - \text{(sum of the solutions)}x + \text{(product of the solutions)}=0}$

${\color{red}x^2 - \left(-\frac{b}{a}\right)x + \left(\frac{c}{a}\right)=0}$

Where did these sum and products come from? Consider the sum of the two solutions obtained from the quadratic formula:

$& r_1 + r_2 = \left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right) + \left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right) \\& r_1 + r_2 = -\frac{b}{2a} + \frac{\sqrt{b^2-4ac}}{2a} -\frac{b}{2a} - \frac{\sqrt{b^2-4ac}}{2a} \\& r_1 + r_2 = -\frac{b}{2a} + \cancel{\frac{\sqrt{b^2-4ac}}{2a}} -\frac{b}{2a} - \cancel{\frac{\sqrt{b^2-4ac}}{2a}} \\& r_1 + r_2 = -\frac{b}{2a} -\frac{b}{2a} \\& r_1 + r_2 = -\frac{2b}{2a} \\& \boxed{r_1 + r_2 = -\frac{b}{a}}$

Now consider the product of the two solutions obtained from the quadratic formula:

$& r_1 \times r_2 = \left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right) \times \left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right) \\& r_1 \times r_2 = \left(-\frac{b}{2a} + \frac{\sqrt{b^2-4ac}}{2a}\right) \times \left(-\frac{b}{2a} - \frac{\sqrt{b^2-4ac}}{2a}\right)$

$& r_1 \times r_2 = \frac{b^2}{4a^2} - \frac{b^2-4ac}{4a^2} \\& r_1 \times r_2 = \frac{b^2-b^2 {\color{red}+}4ac}{4a^2} \\& r_1 \times r_2 = \frac{4ac}{4a^2} \\& \boxed{r_1 \times r_2 = \frac{c}{a}}$

You can use these ideas to determine a quadratic equation with solutions $2+\sqrt{3}$ and $2-\sqrt{3}$ .

• The sum of the solutions is: $2+\sqrt{3}+ 2-\sqrt{3}=4$ .
• The product of the solutions: $(2+\sqrt{3})(2-\sqrt{3})=4-3=1$ .

Therefore, the equation is:

${\color{red}x^2 - \text{(sum of the solutions)}x + \text{(product of the solutions)}=0}$

$x^2-4x+1=0$

#### Example A

Without solving, determine the sum and the product of the solutions for the following quadratic equations:

i) $x^2+3x+2=0$

ii) $3m^2+4m-3=0$

Solution: Remember that the sum of the solutions is $-\frac{b}{a}$ and the product of the solutions is $\frac{c}{a}$ .

i) For $x^2+3x+2=0$ , $a=1, b=3, c=2$ . Therefore, the sum of the solutions is: $-\frac{3}{1}=-3$ . The product of the solutions is $\frac{2}{1}=2$ .

ii) For $3m^2+4m-3=0$ , $a=3, b=4, c=-3$ . Therefore, the sum of the solutions is: $-\frac{4}{3}$ . The product of the solutions is $\frac{-3}{3}=-1$ .

#### Example B

Find a quadratic function with the roots: $3 \pm \sqrt{5}$ .

Solution: The solutions to the quadratic equation are:

$x=3+\sqrt{5}$ and $x=3-\sqrt{5}$

The factors are $(x-(3+\sqrt{5}))$ and $(x-(3-\sqrt{5}))$ . One possible function in factored form is:

$y=\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)$

Multiply and simplify:

$y=x^2-3x+\sqrt{5}x - 3x+9 -3\sqrt{5} - \sqrt{5}x + 3\sqrt{5} - \sqrt{25}$

$& y= x^2-3x + \cancel{\sqrt{5}x} - 3x+9 -\cancel{3\sqrt{5}} -\cancel{\sqrt{5}x} + \cancel{3\sqrt{5}} -{\color{red}5} \\& y=x^2-3x-3x+9-5 \\& \boxed{y=x^2-6x+4}$

Keep in mind that any multiple of the right side of the above function would also have the given roots.

#### Example C

Using the solutions indicated below; determine the quadratic equation by using the sum and the product of the solutions:

$3+2\sqrt{2}$ and $3-2\sqrt{2}$

Solution: The sum of the solutions is $3+2\sqrt{2}+3-2\sqrt{2}=6$ .

The product of the solutions is $(3+2\sqrt{2})(3-2\sqrt{2})=9-8=1$ .

${\color{red}x^2 - \text{(sum of the solutions)}x + \text{(product of the solutions)}=0}$

$x^2-6x+1=0$

#### Concept Problem Revisited

What quadratic function has roots of $2$ and $7$ ? There are multiple functions with these roots. The basic example is $y=(x-2)(x-7)$ which is $y=x^2-9x+14$ . However any function of the form $y=a(x-2)(x-7)$ with $a\ne 0$ would work.

### Vocabulary

The roots of a quadratic function are also the $x$ -intercepts of the function. These are the values for the variable ‘ $x$ ’ that will result in $y = 0$ .
Product of the Roots
Product of the roots is an expression used to find the product of the roots of a given quadratic equation written in general form. The expression used to determine the product of the roots is:
$r_1 \times r_2 = \frac{c}{a}$
Sum of the Roots
Sum of the roots is an expression used to find the sum of the roots of a given quadratic equation written in general form. The expression used to determine the sum of the roots is:
$r_1 + r_2 = -\frac{b}{a}$

### Guided Practice

1. By solving the given equation, find an equation whose solutions are each one less than the solutions to:

$y^2-3y-6=0$

2. Without solving the given equation, find an equation whose solutions are the reciprocals of the solutions to:

$2x^2-3x+5=0$

3. Without solving the given equation, find an equation whose solutions are the negatives of the solutions to:

$m^2-4m+9=0$

1. Determine the solutions of the quadratic equation with the quadratic formula. You should get that the solutions to the quadratic equation are:

$y=\frac{3 + \sqrt{33}}{2} \ or \ y=\frac{3 - \sqrt{33}}{2}$

The solutions of the new equation must be one less than each of the above solutions.

$& y=\frac{3}{2} + \frac{\sqrt{33}}{2} \ or \ y=\frac{3}{2} - \frac{\sqrt{33}}{2} \\& y=\frac{3}{2} + \frac{\sqrt{33}}{2} {\color{red}-1} \ or \ y=\frac{3}{2} - \frac{\sqrt{33}}{2} {\color{red}-1} \\& y=\frac{3}{2} + \frac{\sqrt{33}}{2} {\color{red}-\frac{2}{2}} \ or \ y=\frac{3}{2} - \frac{\sqrt{33}}{2} {\color{red}-\frac{2}{2}} \\& y=\frac{1}{2} + \frac{\sqrt{33}}{2} \ or \ y=\frac{1}{2} - \frac{\sqrt{33}}{2}$

The solutions of the new equation are:

$& y = \frac{1 + \sqrt{33}}{2} \ or \ y = \frac{1 - \sqrt{33}}{2}$

The sum of the solutions is $1$ . The product of the solutions is $-8$ . One possible quadratic equation is $y^2-1y-8=0$ .

2. The sum of the solutions is $\frac{3}{2}$ . The product of the solutions is $\frac{5}{2}$ . The solutions of the new equation must be the reciprocals of the solutions of the original equation. Therefore, the sum of the solutions of the new equation will be:

$& R_1 + R_2 = \frac{1}{r_1} + \frac{1}{r_2} \\& R_1 + R_2 = \frac{1}{r_1} \left({\color{red}\frac{r_2}{r_2}}\right) + \frac{1}{r_2} \left({\color{red}\frac{r_1}{r_1}}\right) \\& R_1 + R_2 = \frac{r_2}{r_1r_2} + \frac{r_1}{r_1r_2} \\& R_1 + R_2 = \frac{r_2 + r_1}{r_1r_2} \\& R_1 + R_2 = \frac{\frac{3}{2}}{\frac{5}{2}} \\& R_1 + R_2 = \frac{3}{2} \times \frac{2}{5} \\& \boxed{R_1 + R_2 = \frac{3}{5}}$

The product of the solutions of the new equation will be:

$& R_1 \times R_2 = \frac{1}{r_1} \times \frac{1}{r_2} \\& R_1 \times R_2 = \frac{1}{r_1r_2} \\& R_1 \times R_2 = \frac{1}{\frac{5}{2}} \\& R_1 \times R_2 = 1 \left(\frac{2}{5}\right) \\& \boxed{R_1 \times R_2 = \frac{2}{5}}$

The new equation is:

$& x^2 - {\color{red}(r_1+r_2)}x + {\color{red}(r_1 \times r_2)} = 0 \\& x^2 - {\color{red}\left(\frac{3}{5}\right)}x + {\color{red}\left(\frac{2}{5}\right)} = 0 \\& 5 \left(x^2 - {\color{red}\left(\frac{3}{5}\right)}x + {\color{red}\left(\frac{2}{5}\right)} = 0 \right) \\& \boxed{5x^2-3x+2=0}$

3. The sum of the solutions is $4$ and the product of the solutions is $9$ . The solutions to the new equation must be negatives of the solutions of the original equation. Therefore, the sum and the product of the new solutions are:

$\boxed{R_1+R_2=-r_1+(-r_2)=-(r_1+r_2)=-4} \ and \ \boxed{R_1 \times R_2 =(-r_1)\times (-r_2)=r_1\times r_2=9}$

The new equation is:

$& m^2 - {\color{red}(r_1 + r_2)}m + {\color{red}(r_1 \times r_2)} = 0 \\& m^2 - {\color{red}(-4)}m + {\color{red}(+9)}= 0 \\& \boxed{m^2 +4m +9 =0}$

### Practice

Without solving, determine the sum and the product of the roots of the following quadratic equations.

1. $2y^2-8y+3=0$
2. $3e^2-6e=4$
3. $0=14-12x+18x^2$
4. $5x^2+6=7x$
5. $2(2x-1)(x+5)=x^2+4$

For the following sums and products of the solutions, state one possible quadratic equation:

1. sum: 4; product: 3
2. sum: 0; product: –16
3. sum: –9; product: –7
4. sum: –6; product: –5
5. sum: $-\frac{2}{3}$ ; product: $\frac{5}{3}$

For the given roots, determine the factors of the quadratic function:

1. $-\frac{3}{2}$ and 5
2. $\frac{1}{4}$ and $\frac{3}{2}$
3. –5 and 3
4. $-\frac{5}{2}$ and $-\frac{4}{3}$
5. $\pm \frac{5}{2}$

For the given roots, determine a potential quadratic function:

1. –2 and –4
2. –3 and $-\frac{1}{3}$
3. $2+ \sqrt{3}$ and $2- \sqrt{3}$
4. $\pm 2\sqrt{5}$
5. $-3 \pm \sqrt{7}$

### Vocabulary Language: English

Function

Function

A function is a relation where there is only one output for every input. In other words, for every value of $x$, there is only one value for $y$.
Roots

Roots

The roots of a function are the values of x that make y equal to zero.