About roots of quadratic trinominals

Question

How can I prove that we can find a point $(p;q)$ on the graph of the trinominal $x^2+bx+c=0$ that has only one root, that makes the equation $x^2+px+q=0$ have one root?

Answer 1

Well, we have that:

$$\text{a}x^2+\text{b}x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\tag1$$

Now, what happens when:

$$\text{b}^2-4\text{a}\text{c}=0\space\Longleftrightarrow\space\sqrt{\text{b}^2-4\text{a}\text{c}}=0\tag2$$

So, we get:

$$x=\frac{-\text{b}\pm\sqrt{0}}{2\text{a}}=\frac{-\text{b}\pm0}{2\text{a}}=-\frac{\text{b}}{2\text{a}}\tag3$$

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In your example we have that:

$$\text{a}=1,\text{b}=\text{p},\text{c}=\text{q}\tag4$$

So we get one root when:

$$\text{p}^2-4\cdot1\cdot\text{q}=\text{p}^2-4\text{q}=0\space\Longleftrightarrow\space\color{red}{\frac{\text{p}^2}{\text{q}}=4}\tag5$$