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Roots to Determine a Quadratic Function

Compose quadratic functions from given roots

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Roots to Determine a Quadratic Function

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

Watch This

James Sousa: Find a Quadratic Function with Fractional Real Zeros

Guidance

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

\begin{align*}y&=(x-2)(x-5)\\ y&=x^2-7x+10\end{align*}

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

\begin{align*}y&=2(x-2)(x-5)\\ y&=2x^2-14x+20\end{align*}

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 \begin{align*}r_1\end{align*} and \begin{align*}r_2\end{align*} are the solutions to the quadratic equation \begin{align*}ax^2+bx+c=0\end{align*}, then

\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:

\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:

\begin{align*}& 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}}\end{align*}

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

\begin{align*}& 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)\end{align*}

\begin{align*}& 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}}\end{align*}

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:

\begin{align*}y=\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\end{align*}

Multiply and simplify:

\begin{align*}y=x^2-3x+\sqrt{5}x - 3x+9 -3\sqrt{5} - \sqrt{5}x + 3\sqrt{5} - \sqrt{25} \end{align*}

\begin{align*}& 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}\end{align*}

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

Roots of a Quadratic Function
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:

\begin{align*}y^2-3y-6=0\end{align*}

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

\begin{align*}2x^2-3x+5=0\end{align*}

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

\begin{align*}m^2-4m+9=0\end{align*}

Answers:

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

\begin{align*}y=\frac{3 + \sqrt{33}}{2} \ or \ y=\frac{3 - \sqrt{33}}{2}\end{align*}

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

\begin{align*}& 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}\end{align*}

The solutions of the new equation are:

\begin{align*}& y = \frac{1 + \sqrt{33}}{2} \ or \ y = \frac{1 - \sqrt{33}}{2}\end{align*}

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:

\begin{align*}& 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}}\end{align*}

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

\begin{align*}& 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}}\end{align*}

The new equation is:

\begin{align*}& 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}\end{align*}

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:

\begin{align*}\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}\end{align*}

The new equation is:

\begin{align*}& 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}\end{align*}

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

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.

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