Question

$\rightarrow ax^2+bx+c$

if Quadratic function f(x) gives value $\frac{-(b^2-4ac)}{8a}$ on the input $x_1$ and $x_2$ Respectively, iff $x_1+x_2=12$

Then value of $\frac{-b}{8a}$

also given: $D>0$

Answer 1

3/2

Answer 2

The quadratic function is given by $f(x) = ax^2+bx+c$. We are given that $f(x_1) = \frac{-(b^2-4ac)}{8a}$ and $f(x_2) = \frac{-(b^2-4ac)}{8a}$. Let $D = b^2-4ac$. So, $f(x_1) = \frac{-D}{8a}$ and $f(x_2) = \frac{-D}{8a}$. This means that $x_1$ and $x_2$ are the roots of the equation $f(x) = \frac{-D}{8a}$. Substituting $f(x) = ax^2+bx+c$, we get: $ax^2+bx+c = \frac{-D}{8a}$ $ax^2+bx+c - \left(\frac{-D}{8a}\right) = 0$ $ax^2+bx+c + \frac{D}{8a} = 0$ $ax^2+bx + \frac{8ac+D}{8a} = 0$ Since $D = b^2-4ac$, we have $ax^2+bx + \frac{8ac+(b^2-4ac)}{8a} = 0$ $ax^2+bx + \frac{b^2+4ac}{8a} = 0$

Let $x_1$ and $x_2$ be the roots of this quadratic equation. According to Vieta's formulas, the sum of the roots of a quadratic equation $Ax^2+Bx+C=0$ is given by $-\frac{B}{A}$. In our case, $A=a$, $B=b$, and $C=\frac{b^2+4ac}{8a}$. So, the sum of the roots $x_1+x_2 = \frac{-b}{a}$.

We are given in the problem that $x_1+x_2=12$. Therefore, we can equate the two expressions for the sum of the roots: $\frac{-b}{a} = 12$

The question asks for the value of $\frac{-b}{8a}$. We can express $\frac{-b}{8a}$ in terms of $\frac{-b}{a}$: $\frac{-b}{8a} = \frac{1}{8} \times \left(\frac{-b}{a}\right)$ Substitute the value $\frac{-b}{a}=12$: $\frac{-b}{8a} = \frac{1}{8} \times 12$ $\frac{-b}{8a} = \frac{12}{8}$ $\frac{-b}{8a} = \frac{3}{2}$

The condition $D>0$ ensures that the value $\frac{-D}{8a}$ is distinct from the vertex value $\frac{-D}{4a}$ (since $a \ne 0$), and thus there are indeed two distinct values $x_1$ and $x_2$ for which $f(x)$ takes this value.

Explanation of the solution:

1. Identify $x_1$ and $x_2$ as roots of the equation $f(x) = \frac{-(b^2-4ac)}{8a}$.
2. Rewrite the equation as $ax^2+bx+(c - \frac{-(b^2-4ac)}{8a}) = 0$.
3. Simplify the constant term: $c + \frac{b^2-4ac}{8a} = \frac{8ac+b^2-4ac}{8a} = \frac{b^2+4ac}{8a}$.
4. The equation becomes $ax^2+bx+\frac{b^2+4ac}{8a} = 0$.
5. Use Vieta's formulas for the sum of roots: $x_1+x_2 = -\frac{b}{a}$.
6. Given $x_1+x_2=12$, so equate: $-\frac{b}{a}=12$.
7. Calculate the required value $\frac{-b}{8a} = \frac{1}{8} \times \left(-\frac{b}{a}\right) = \frac{1}{8} \times 12 = \frac{3}{2}$.