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If $a$, $b$, $c$ are real numbers such that $2a + 3b + 6c = 0$, prove that $ax^2+bx +c=0$ has a solution in the interval $[0, 1]$.

This should use high school maths or a little bit more then that.

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    Ok i just don't understand why you chose $f(1/2)$. – 2017-01-27
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    If you want to respond the answer, do comment under the answer, this ensure the one who answered being informed. No one can review all questions answered all the time. – 2017-01-27

1 Answers 1

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\begin{align*} 0 &= 2a+3b+6c \\ f(x) &= ax^2+bx+c \\ f(0) &= c \\ f\left( \frac{1}{2} \right) &= \frac{a}{4}+\frac{b}{2}+c \\ &= -\frac{a}{12} \\ f(1) &= a+b+c \\ &= \frac{a}{3}-c \end{align*}

Case I: $ac \ge 0$

$$f(0) f\left( \frac{1}{2} \right) = -\frac{ac}{12} \le 0$$

  • $\exists x\in \left[ 0, \dfrac{1}{2} \right]$ such that $f(x)=0$.

Case II: $ac \le 0$

$$f\left( \frac{1}{2} \right) f(1)= -\frac{a^2}{36}+\frac{ac}{12} \le 0$$

  • $\exists x\in \left[ \dfrac{1}{2}, 1 \right]$ such that $f(x)=0$.

Combining, $\exists x \in [0,1]$ such that $f(x)=0$

N.B.

When $0 <3ac two such roots, namely $\alpha \in \left( 0, \frac{1}{2} \right)$ and $\beta \in \left( \frac{1}{2}, 1 \right)$.

Updates

The value of $x=\dfrac{1}{2}$ can be inspired by plotting a family of curves by varying either $b$ or $c$. You can see a fixed point at $x=\dfrac{1}{2}$ by varying $c$ below.

enter image description here

Alternatively, $(x,y)=\left(\dfrac{1}{2},-\dfrac{a}{12} \right)$ is a solution of

$$ax^2-\left( \frac{2a}{3} + 2c \right)x+c-y= \frac{\partial}{\partial c} \left[ ax^2-\left( \frac{2a}{3} + 2c \right)x+c-y \right]=0$$

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    Nice Solution Ng Chung Tak, But i did not understand your alternate solution , specially $\displaystyle \frac{\partial}{\partial c} \left[ ax^2-\left( \frac{2a}{3} + 2c \right)x+c-y \right]=0$ – 2017-01-30
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    That's from theory of [**envolope**](https://en.wikipedia.org/wiki/Envelope_(mathematics)). If it's beyond your level, just ignore it. – 2017-01-30