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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\begin{align*}2\end{align*} and 7\begin{align*}7\end{align*}? Does more than one function have these roots?

### Guidance

If x=2\begin{align*}x=2\end{align*} and x=5\begin{align*}x=5\end{align*} are roots of a quadratic function, then (x2)\begin{align*}(x - 2)\end{align*} and (x5)\begin{align*}(x-5)\end{align*} must have been factors of the quadratic equation. Therefore, a quadratic function with roots 2\begin{align*}2\end{align*} and 5\begin{align*}5\end{align*} is:

yy=(x2)(x5)=x27x+10

Note that there are many other functions with roots of 2\begin{align*}2\end{align*} and 5\begin{align*}5\end{align*}. These functions will all be multiples of the function above. The function below would also work:

yy=2(x2)(x5)=2x214x+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 r1\begin{align*}r_1\end{align*} and r2\begin{align*}r_2\end{align*} are the solutions to the quadratic equation ax2+bx+c=0\begin{align*}ax^2+bx+c=0\end{align*}, then

r1+r2=ba and r1×r2=ca\begin{align*}{\color{red}r_1 + r_2 = -\frac{b}{a} \ and \ r_1 \times r_2 = \frac{c}{a}}\end{align*}.

The quadratic can be rewritten as:

x2(sum of the solutions)x+(product of the solutions)=0\begin{align*}{\color{red}x^2 - \text{(sum of the solutions)}x + \text{(product of the solutions)}=0}\end{align*}

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

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

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

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

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

Therefore, the equation is:

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

\begin{align*}x^2-4x+1=0\end{align*}

#### Example A

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

i) \begin{align*}x^2+3x+2=0\end{align*}

ii) \begin{align*}3m^2+4m-3=0\end{align*}

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

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

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

#### Example B

Find a quadratic function with the roots: \begin{align*}3 \pm \sqrt{5}\end{align*}.

Solution: The solutions to the quadratic equation are:

\begin{align*}x=3+\sqrt{5}\end{align*} and \begin{align*}x=3-\sqrt{5}\end{align*}

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

Multiply and simplify:

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:

\begin{align*}3+2\sqrt{2}\end{align*} and \begin{align*}3-2\sqrt{2}\end{align*}

Solution: The sum of the solutions is \begin{align*}3+2\sqrt{2}+3-2\sqrt{2}=6\end{align*}.

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

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

\begin{align*}x^2-6x+1=0\end{align*}

#### Concept Problem Revisited

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

### Vocabulary

The roots of a quadratic function are also the \begin{align*}x\end{align*}-intercepts of the function. These are the values for the variable ‘\begin{align*}x\end{align*}’ that will result in \begin{align*}y = 0\end{align*}.
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:
\begin{align*}r_1 \times r_2 = \frac{c}{a}\end{align*}
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:
\begin{align*}r_1 + r_2 = -\frac{b}{a}\end{align*}

### Guided Practice

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

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

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

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

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

The solutions of the new equation are:

The sum of the solutions is \begin{align*}1\end{align*}. The product of the solutions is \begin{align*}-8\end{align*}. One possible quadratic equation is \begin{align*}y^2-1y-8=0\end{align*}.

2. The sum of the solutions is \begin{align*}\frac{3}{2}\end{align*}. The product of the solutions is \begin{align*}\frac{5}{2}\end{align*}. 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:

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

The new equation is:

3. The sum of the solutions is \begin{align*}4\end{align*} and the product of the solutions is \begin{align*}9\end{align*}. 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:

The new equation is:

### Practice

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

1. \begin{align*}2y^2-8y+3=0\end{align*}
2. \begin{align*}3e^2-6e=4\end{align*}
3. \begin{align*}0=14-12x+18x^2\end{align*}
4. \begin{align*}5x^2+6=7x\end{align*}
5. \begin{align*}2(2x-1)(x+5)=x^2+4\end{align*}

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: \begin{align*}-\frac{2}{3}\end{align*}; product: \begin{align*}\frac{5}{3}\end{align*}

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

1. \begin{align*}-\frac{3}{2}\end{align*} and 5
2. \begin{align*}\frac{1}{4}\end{align*} and \begin{align*}\frac{3}{2}\end{align*}
3. –5 and 3
4. \begin{align*}-\frac{5}{2}\end{align*} and \begin{align*}-\frac{4}{3}\end{align*}
5. \begin{align*}\pm \frac{5}{2}\end{align*}

For the given roots, determine a potential quadratic function:

1. –2 and –4
2. –3 and \begin{align*}-\frac{1}{3}\end{align*}
3. \begin{align*}2+ \sqrt{3}\end{align*} and \begin{align*}2- \sqrt{3}\end{align*}
4. \begin{align*}\pm 2\sqrt{5}\end{align*}
5. \begin{align*}-3 \pm \sqrt{7}\end{align*}

### 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.