Construct a quadratic equation that has as its solutions $$x_1=\dfrac{1}{3}$$, $$x_2=-\dfrac{2}{5}$$.
Doing the corresponding product
$$$(x-\dfrac{1}{3})\cdot(x+\dfrac{2}{5})=x^2+\dfrac{1}{15}x-\dfrac{2}{15}$$$ We can remove denominators by multiplying by $$15$$, so we have the equation $$15x^2+x-2=0$$.
$$15x^2+x-2=0$$
The sum of two numbers is $$9$$ and its product $$20$$. Find the values of these numbers.
Applying the formula $$x^2-sx+p=0$$, in this case $$s = 9$$ and $$p = 20$$.
$$$x^2-9x+20=0$$$ $$$\displaystyle x=\frac{9 \pm \sqrt{81-80}}{2}= \frac{9 \pm 1}{2}=\left \{\begin{matrix} x_1=5 \\ x_2=4\end{matrix}\right.$$$ So, the numbers we are looking for are $$5$$ and $$4$$.
$$5$$ and $$4$$
Construct a quadratic equation with discriminant equal to zero and with one of the solutions equal to $$-6$$.
If $$D = 0$$ means that the equation has a double root and, as this has to be $$-6$$, we will have $$$(x-6)\cdot(x-6)=x^2-12x+36$$$
$$x^2-12x+36$$
A school playground is $$600$$ square meters. We needed $$100$$ meters of fencing to fence it. What are the dimensions of the playground?
Let's call $$a$$ and $$b$$ the sides of the rectangle. We know that $$a\cdot b = 600$$ and $$2a + 2b = 100$$, or, the same, $$a + b = 50$$.
Applying the formula $$x^2-sx+p=0$$ we will find that the corresponding quadratic equation is: $$x^2-50x+600=0$$. $$$\displaystyle x=\frac{50 \pm \sqrt{50^2-4\cdot600}}{2}= \frac{50 \pm \sqrt{100}}{2}=\frac{50 \pm 10}{2}=\left\{\begin{matrix} x_1=30 \\ x_2=20\end{matrix}\right.$$$ Then, the playground will be $$30$$ meters long and $$20$$ wide.
$$a = 30$$m and $$b = 20$$m