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

Development:

Doing the corresponding product

$$(x-{\displaystyle \frac{1}{3}})\cdot (x+{\displaystyle \frac{2}{5}})=x^2+{\displaystyle \frac{1}{15}}x-{\displaystyle \frac{2}{15}}$$ We can remove denominators by multiplying by $15$, so we have the equation $15x^2+x-2=0$.

Solution:

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

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Development:

Applying the formula $x^2-sx+p=0$, in this case $s=9$ and $p=20$.

$$x^2-9x+20=0$$ $${\displaystyle x=\frac{9\pm \sqrt{81-80}}{2}=\frac{9\pm 1}{2}=\{\begin{array}{c}{x}_1=5\\ x_2=4\end{array}}$$ So, the numbers we are looking for are $5$ and $4$.

Solution:

$5$ and $4$

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Development:

If $D=0$ means that the equation has a double root and, as this has to be $-6$, we will have $$(x-6)\cdot (x-6)=x^2-12x+36$$

Solution:

$x^2-12x+36$

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Development:

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: $x^2-50x+600=0$. $${\displaystyle x=\frac{50\pm \sqrt{{50}^2-4\cdot 600}}{2}=\frac{50\pm \sqrt{100}}{2}=\frac{50\pm 10}{2}=\{\begin{array}{c}{x}_1=30\\ x_2=20\end{array}}$$ Then, the playground will be $30$ meters long and $20$ wide.

Solution:

$a=30$m and $b=20$m

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