Value of $a$ for which at least one root is greater than $0$

Question

Find sum of all integral values of $a$ in $[1,100]$ for which the equation $x^2-(a-5)x+(a-15/4)=0$ has at least one root greater than zero.

I used the condition that discriminant must be greater than or equal to zero and obtained that $a \in [1,4] \cup [10,100]$, but I am not able to visualize the condition for 'at least one root greater than zero'. Please provide some insight.

Answer 1

Hint see the equations $x^2-3x+2=0$ and $x^2-x+2=0$ . You will see that necessary conditions are $a-5>0,a-\frac{15}{4}>0$

Answer 2

Hint: If the roots of a quadratic are $\alpha$ and $\beta$ then the quadratic is of the form $x^2 - (\alpha + \beta)x + \alpha \beta$.

Now try resolving your question about one of the two roots being greater than $0$ to what that means for the sum and product of them.

Answer 3

The roots of given polynomial are $$ \frac{b\pm\sqrt{b^2-4c}}{2} $$ where $b = a-5$ and $c = a-\frac{15}{4}$.

Now:

• If $b>0$, then $\frac{b+\sqrt{b^2-4c}}{2}>0$.
• If $b\le 0$ but $c<0$, then $b^2-4c>b^2\implies \sqrt{b^2-4c}>|b|$. What can you conclude?
• Otherwise, $b\le 0$ and $c\ge 0$, and so $\sqrt{b^2-4c}\le |b|$. What can you conclude?