Exercise 1.16 [bomc]

[nafistiham]

The polynomial $ax^{2}+bx+c$ satisfies : $a>0$ .$a+b+c\geq0.$ $a-b+c\geq0.a-c\geq0$ and $b^{2}-4ac\geq 0.$ prove that the roots are real and that they belong to the interval $-1\leq x\leq 1$

[sourav das]

Hint

If $\alpha$ and $\beta$ are 2 roots of $ax^2 + bx + c = 0$ then $\alpha + \beta = - \frac {b}{a}$ and $ \alpha \beta = \frac {c}{a}$

[*Mahi*]

We have to prove that $| \frac {-b \pm \sqrt {b^2-4ac} } {2a} | \leq 1$
Or $| {-b \pm \sqrt {b^2-4ac} } | \leq 2a$
As $b^2-4ac \leq (a-c)^2$ so all we have to prove is $2a \geq |b| + |a-c|$ or $b-c \leq a$, which is true indeed.

[sourav das]

Solution

If $\alpha$ and $\beta$ are 2 roots then
$\alpha +\beta = - \frac{b}{a}$
$\alpha \beta = \frac{c}{a}\leq 1$
$\left ( \alpha +1 \right )\left ( \beta +1 \right )= \frac{a-b+c}{a}\geq 0$
$\left ( \alpha -1 \right )\left ( \beta -1 \right )= \frac{a+b+c}{a}\geq 0
$
And these three inequality holds when ($\alpha , \beta$)$\leq 1$

[nafistiham]

cool problem with interesting solutions. actually i was kinda scared seeing all those satisfactions.so couldn't think in these ways.

[sm.joty]

In this problem \[-1\leqslant{} x \]
\[\Leftrightarrow-1\leqslant{} \frac{-b\pm\sqrt{b^2-4ac}}{2a} \]
\[\Leftrightarrow (b-2a)^2\leqslant{} b^2-4ac \]
\[\Leftrightarrow 4a^2-4ab\leqslant{} -4ac \]
\[\Leftrightarrow a-b+c\leqslant{} 0\]
But it's a contradiction!!!
HOW COULD IT POSSIBLE ?
is there any mistake gays..

[*Mahi*]

Squaring is not always valid in inequalities. It is true only when both sides of the inequality are positive . Easy contradictions can be shown like $2>-3$ and $-3 < -2$.

[FahimFerdous]

Yeah, squaring in case of inequalities is not valid always.

[nafistiham]

so in which cases we can say squaring is valid or not?
i mean what the properties are?

[sourav das]

Actually only when absolute value of a is greater than absolute value of b ; we can square both side.