Find $a,b,c$ and intersection of $y = ax^2 + bx + c$ with $x$-axis

Question

Known: max at $x = 1, y = 16$, intersect $x$-axis with the length of the section connecting the two points is $8$.

What I tried is this:

$f'(x) = 2ax + b$, max at $x = 1, so -b = 2a, y = 16 \rightarrow a + b + c = 16$ and $x_2 - x_1 = \frac{\sqrt{D}}{a}$

How do I continue?

Answer 1

You have: $$b=-2a$$ $$a+b+c=16$$ and $$8=\frac{\sqrt{b^2-4ac}}{a}$$ Eliminating $b$ in the second and third equations with the help of the first, we get $$c-a=16$$ and $$a(15a+c)=0$$ Can you finish?

Answer 2

This is what you are missing. By Vieta's formulas:

$$(x_2 - x_1)^2 = (x_2 + x_1)^2 - 4x_1x_2 = \frac{b^2}{a^2} - 4\frac{c}{a}$$ $$\Rightarrow 64 = \frac{b^2}{a^2} - 4\frac{c}{a}$$ $$\Rightarrow 64a^2 = b^2 - 4ac$$

Since $-b = 2a$, $a-2a+c=16 \Rightarrow c = 16+a$:

$$64a^2 = 4a^2 - 4ac$$ $$\Rightarrow 60a^2 = -4a(16+a)$$ $$\Rightarrow 60a^2 + 64a + 4a^2 = 0$$

Answer 3

For the roots of quadratic equation

half sum=1=HS; sum =2

difference =8; half difference =4=HD

Product= HS^2-HD^2=1^2-4^2=-15

The quadratic has the form with sum and product of roots

$$ y= A(x^2 -x(2)-15)= A(x-5)(x+3)$$

The roots are ( intersection with x-axis) $(5m-3)$.

At maximum x=1, y= 16 as given. Plug this in

and evaluate the constant coefficient: $A=-1$