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.

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

$$\displaystyle x=\frac{-2 \pm \sqrt{4+12}}{2}=\frac{-2 \pm 4}{2}=\left\{\begin{matrix} x_1=1\\ x_2=-3 \end{matrix}\right.$$

Now let's look at what happens when we do the product $(x-x_1) \cdot (x-x_2)$

$$(x-1) \cdot (x+3)=x^2-x+3x-3=x^2+2x-3$$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".

If the solutions of the equation are $x_1=4,x_2=2$ the corresponding quadratic equation is:

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

If the roots of the equation are $x_1=-1, x_2=-5$ the corresponding quadratic equation is:

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

If the solutions of the equation are $x_1=3, x_2= \displaystyle -\frac{2}{3}$ the corresponding quadratic equation is:

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

If the roots of the equation are $x_1=0, x_2=16$ the corresponding quadratic equation is:

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

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:

$$(x-x_1)\cdot (x-x_2)=x^2-x_1x-x_2x+x_1x_2=x^2-(x_1+x_2)x+x_1x_2$$

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:

$$x^2-sx+p=0$$

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:

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

This method is faster than doing the product of roots.

Let's see some other examples:

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

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

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

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

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.

If we want to get the equation $x^2-5x+6=0$ 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:

$$\displaystyle x=\frac{5 \pm \sqrt{25-24}}{2}= \frac{5 \pm 1}{2}= \left\{ \begin{matrix} x_1=3 \\ x_2=2\end{matrix} \right.$$

With these same values we can approach it geometrically.

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

The perimeter of a rectangle is the sum of all its sides, then it is $a+a+b+b = 2a + 2b= 2(a+b) = 10$, 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 $x^2-5x+6=0$.

And the solution is:

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

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