 # Quadratic Equations From Their Solutions From solutions to factors to final equations (10.3)

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Quadratic Equations From Their Solutions From solutions to factors to final equations (10.3)

POD Solve the following equations.

POD Solve the following equations. We moved from equation to solution here.

Today we start with solutions Then we determine the factors from those solutions. Then we multiply the factors to find the final equation. It’s the opposite direction from the POD.

Try it 1. The solutions for this quadratic equation are 5 and -7, and the leading coefficient is 1. Find the factors. Give the equation.

Try it 1. The solutions for this quadratic equation are 5 and -7, and the leading coefficient is 1. Find the factors. (x - 5) and (x + 7) Give the equation. y = (x – 5)(x + 7) y = x 2 + 2x - 35

Try it 2. The solutions are 5/3 and -2. Give an equation with a leading coefficient of one. Give an equation with integer coefficients.

Try it 2. The solutions are 5/3 and -2. Give an equation with a leading coefficient of one. The factors are (x – 5/3) and (x + 2). The equation is y = (x – 5/3)(x + 2) y = x 2 + x/3 – 10/3.

Try it 2. The solutions are 5/3 and -2. Give an equation with integer coefficients. All we have to do is multiply every term by 3. y = 3x 2 + x – 10. Although it’s a different parabola, it has the same zeros.

Try it 3. The solutions are 1/4 and 2/3. Find an equation with integer coefficients.

Try it 3. The solutions are 1/4 and 2/3. Find an equation with integer coefficients. The factors are (x – 1/4) and (x – 2/3). The equation is y = x 2 – 11x/12 + 1/6. With integer coefficients it would be y = 12x 2 – 11x + 2. How would you check it?

Try it 3. With integer coefficients it would be y = 12x 2 – 11x + 2. Check it using the quadratic formula.

Try it 4. The solutions are 2+i and 2-i.

Try it 4. The solutions are 2+i and 2-i. The factors will be (x-(2+i)) and (x-(2-i)).

Try it 4. FOILing will be useful here. How would you check your answer?

The Rule Imaginary solutions come in complex conjugate pairs. So, if 2 + 3i is a solution, what must another solution be? What is an equation?

The Rule Imaginary solutions come in complex conjugate pairs. So, if 2 + 3i is a solution, what must another solution be?2 - 3i What is an equation? y = x 2 - 4x + 13

Factoring with imaginary roots The process is the same, whether the roots are real or imaginary: 1. With imaginary roots, just determine the complex conjugates. 2. Once you have roots, draft the factors. 3. Multiply the factors.

Reality check Real roots for a parabola (or any function) mean that the graph of the function crosses the x-axis at that point. In other words, they are _____________

Reality check Real roots for a parabola (or any function) mean that the graph of the function crosses the x-axis at that point. In other words, they are x-intercepts.

Reality check Imaginary roots for a parabola mean that the graph of the function does not cross the x-axis at all. There are no x-intercepts.

Try it 4. Make your own. Start with a complex number solution. Find the complex conjugate that must be the other solution. Then find the factors and the equation. Everyone use a small white board!