Quadratic equation given its solutions or the sum and product of roots

Construction of a quadratic equation from its solutions

We are going to see now the way we can construct a quadratic equation when the solutions are known.

Example

The solutions of the equation $x^{2} + 2 x - 3 = 0$ are:

$x = \frac{- 2 \pm \sqrt{4 + 12}}{2} = \frac{- 2 \pm 4}{2} = {\begin{matrix} x_{1} = 1 \\ x_{2} = - 3 \end{matrix}$

Now let's look at what happens when we do the product $(x - x_{1}) \cdot (x - x_{2})$

$(x - 1) \cdot (x + 3) = x^{2} - x + 3 x - 3 = x^{2} + 2 x - 3$ We have returned to the original equation.

So "the product of $x$ minus a root multiplied by $x$ minus the other root is equal to the quadratic equation that has these roots as a solution".

Example

If the solutions of the equation are $x_{1} = 4, x_{2} = 2$ the corresponding quadratic equation is:

$(x - 4) (x - 2) = x^{2} - 6 x + 8 = 0$

Example

If the roots of the equation are $x_{1} = - 2, x_{2} = - 5$ the corresponding quadratic equation is:

$(x + 1) (x + 5) = x^{2} + 6 x + 5 = 0$

Example

If the solutions of the equation are $x_{1} = 3, x_{2} = - \frac{2}{3}$ the corresponding quadratic equation is:

$(x - 3) (x + \frac{2}{3}) = x^{2} - \frac{7}{3} x - 2 = 0$

Example

If the roots of the equation are $x_{1} = 0, x_{2} = 16$ the corresponding quadratic equation is:

$(x - 0) (x - 16) = x^{2} + 16 x = 0$

Reconstruction of the quadratic equation from the sum and product of roots

We know that $(x - x_{1}) \cdot (x - x_{2})$ leads to the equation that has $x_{1}, x_{2}$ as its solutions. If we do the product:

$(x - x_{1}) \cdot (x - x_{2}) = x^{2} - x_{1} x - x_{2} x + x_{1} x_{2} = x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2}$

an expression in which appear the sum and the product of the roots, let's call them $s$ and $p$ .

$s = x_{1} + x_{2} p = x_{1} \cdot x_{2}$

So the quadratic equation is:

$x^{2} - s x + p = 0$

Example

Write a quadratic equation knowing that the sum of its roots is $5$ and its product $6$ .

We know that $s = 5, p = 6$ , then the equation will be:

$x^{2} - 5 x + 6 = 0$

This method is faster than doing the product of roots.

Let's see some other examples:

Example

The quadratic equation that has solutions $4$ and $9$ is:

$x^{2} - 13 x + 36 = 0$

Example

The quadratic equation that has solutions $- 3$ and $- 5$ is:

$x^{2} + 8 x + 15 = 0$

Let's say it is not easy to lay out an exercise that ends with a quadratic equation. The easiest way would be writing literally what the equation says.

Example

If we want to get the equation $x^{2} - 5 x + 6 = 0$ as a solution to a problem, we can formulate a statement like: If we raise an amount to the square and we subtract $5$ times this amount the result is $- 6$ . What is the value of that amount?

The following statement is clearly much more interesting: "Find two numbers knowing that their sum is $5$ and their product is $6$ " , a statement that ends with the same equation and whose solutions can be found solving the proposed equation:

$x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} = {\begin{matrix} x_{1} = 3 \\ x_{2} = 2 \end{matrix}$

With these same values we can approach it geometrically.

Example

We know that the perimeter of a rectangle is $10$ and its area $6$ . Calculate the sides of this rectangle.

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The perimeter of a rectangle is the sum of all its sides, then it is $a + a + b + b = 2 a + 2 b = 2 (a + b) = 10$ , that is, $a + b = 5$

On the other side, the area of the rectangle is $a \cdot b = 6$ .

Then, to solve this problem we have to solve a quadratic equation in which the sum of its roots is $5$ and its product $6$ . This equation is $x^{2} - 5 x + 6 = 0$ .

And the solution is:

$x = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}$

Then, the sides of the rectangle will be, $a = 2$ and $b = 3$