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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?

Watch This

James Sousa: Find a Quadratic Function with Fractional Real Zeros

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

Roots of a Quadratic Function
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

Answers:

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}

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