Quadratic equations and probability

Question

The inequality:

4p^2-17p+4>0

Solving using quadratic equation:

(−(−17)±√(−17)^2−4⋅4⋅4)/8 =(12±√225)/8

I realize why p = 4 or p = 1/4, and in this case p represents and probability so the solution is 1/4 but how do I know if p is < or > than 1/4.

Thanks.

Answer 1

$4p^2 - 17p + 4 = (4p - 1)(p - 4) > 0$.

You already know that $0 \le p \le 1$ since it is a probability, so that means $p < 4$. So $p - 4 < 0$. This means: $4p - 1 < 0$, and so $p < \frac{1}{4}$

Answer 2

4p^2 - 17p +4 >0 First solve for the critical values 4p^2 -17p + 4 = 0 p= 4, p= 1/4

Draw a number line , and mark 4 and 1/4 , if it is a function greater than 0 which is in this case, draw an arrow from 4 away from 1/4 which is p> 4 , and same for 1/4 u draw a line away from 4 which is p < 1/4. So the solution is p>4 , p<1/4.

Now imagine if it was 4p^2 - 17p + 4<0 Less than zero ^^^ then you draw a arrow inwards from 1/4 towards 4 and same thing for 4 u draw towards 1/4 , which gives the solution region as 1/4

Answer 3

Completing the square gives:

$(2p-\frac{17}{4})^{2}>(\frac{17}{4})^{2}-4$

Taking the square root of both sides gives:

$$|2p-\frac{17}{4}|>\frac{15}{4} \\\iff 2p-\frac{17}{4}>\frac{15}{4} \text{ or } 2p-\frac{17}{4}<-\frac{15}{4}\\ \iff p>4\text{ or } p<\frac{1}{4}$$

Answer 4

$$4p^2-17p+4=(4p-1)(p-4)>0$$ $$\iff(4p-1>0\&p-4>0)\text{or}(4p-1<0\&p-4<0)$$ $$\iff(p>1/4\&p>4)\text{or}(p<1/4\&p<4)$$ $$\iff(p>4)\text{or}(p<1/4)$$ $$\iff p\in(-\infty,1/4)\cup(4,\infty)$$ since $p$ is probability $0\leq p\leq1$ after combining these results finally $0\leq p<1/4$