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?
Asked
Active
Viewed 25 times
1 Answers
0
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:
$$x=\frac{-\text{b}\pm\sqrt{0}}{2\text{a}}=\frac{-\text{b}\pm0}{2\text{a}}=-\frac{\text{b}}{2\text{a}}\tag3$$
In your example we have that:
$$\text{a}=1,\text{b}=\text{p},\text{c}=\text{q}\tag4$$
So we get one root when:
$$\text{p}^2-4\cdot1\cdot\text{q}=\text{p}^2-4\text{q}=0\space\Longleftrightarrow\space\color{red}{\frac{\text{p}^2}{\text{q}}=4}\tag5$$
Jan Eerland
- 28,671
-
And how can I prove that we will be able to find points (p;q) on the graph of x^2+bx+c=0 such as p^2 = 4q? – student28 Feb 26 '17 at 13:06
-
@student28 I've proved it already, because there will be an infinite number of points where that condition is true!! – Jan Eerland Feb 26 '17 at 13:15