What is the interval for which $a$ exists?

Question

Question :- For $x$ belonging to $[3,10]$ the following expression $x^2 + 2ax + 10 - 3a$ is always greater than $0$. What is the interval for which $a$ exists?
My Attempt :- I differentiated the quadratic expression and found that the abscissa of the vertex of the parabola of the given quadratic expression lies at $x = -a$.
After this I really do not know what to do. Though I was thinking that if this expression is always greater than $0$ than it's discriminant needs to be less than $0$. Maybe I can find the interval for which $a$ exists in this way. This discriminant method would have definitely worked if $x$ just needed to be a real number. But when the $x$ is limited to $[3,10]$, I am not sure about this method.
Please clarify my doubt.

Answer 1

Let's note $f(x)=x^2+2ax+10-3a$.

As you correctly established, "I differentiated the quadratic expression and found that the abscissa of the vertex of the parabola of the given quadratic expression lies at $x=−a$.", and the parabola is U-shaped, hence the minimum value of $f$ on $\mathbb{R}$ is $f(-a)$.

Your exercise is just to find the minimum value of $f$ on $[3,10]$ (and compare it to $0$), not on $\mathbb{R}$, so the answer will depend whether $-a$ is in that interval, on its left or on its right.

That means you can consider three cases:

• if $-a\in[3,10]$, i.e. $a\in[-10,-3]$, then the minimum value of your expression is $f(-a)$, you just have to solve $f(-a)>0$ for $a$.

• if $-a<3$, i.e. $a>-3$, then the minimum value of your expression is $f(3)$, you just have to calculate $f(3)$ and solve $f(3)>0$ for $a$.

• if $-a>10$, i.e. $a<-10$, then the minimum value of your expression is $f(10)$, you just have to calculate $f(10)$ and solve $f(10)>0$ for $a$.

Answer 2

Note that :$$f(3)>0\\f(10)>0\\f(-a)>0$$ all them must be positive simultaneously $$f(x)=(x+a)^2-a^2+10-3a\to \text{min is in } x=-a$$so $$6a+10>0\\20a+80>0\\-a^2+10-3a>0$$ $$a>\frac{-10}{6}\\a>\frac{-80}{20}\\(a+5)(a-2)<0 \to a \in (-5,2)$$can you go on ?