If-th-roots-of-the-equation-x-2-2ax-b-0-are-real-and-disinct-and-they-differ-by-at-most-2m-then-b-lies-in-the-interval-

Question Number 21519 by ram1234 last updated on 26/Sep/17

If th roots of the equation x^{2} + 2 a x + b = 0

are real and disinct and they differ by

at most 2 m, then b lies in the interval

Answered by mrW1 last updated on 26/Sep/17

x^{2} + 2 ax + b = 0

x^{2} + 2 ax + a^{2} - (a^{2} - b) = 0

{(x + a)}^{2} - {(\sqrt{a^{2} - b})}^{2} = 0

(x + a + \sqrt{a^{2} - b}) (x + a - \sqrt{a^{2} - b}) = 0

x_{1} = - a + \sqrt{a^{2} - b}

x_{2} = - a - \sqrt{a^{2} - b}

x_{1} \neq x_{2}

\Rightarrow b \neq a^{2}

x_{1} - x_{2} = 2 \sqrt{a^{2} - b} ⩽ 2 m

\Rightarrow \sqrt{a^{2} - b} ⩽ m

\Rightarrow - m^{2} ⩽ a^{2} - b ⩽ m^{2}

\Rightarrow a^{2} - m^{2} ⩽ b ⩽ a^{2} + m^{2} \cup b \neq a^{2}

Commented by Joel577 last updated on 26/Sep/17

{What}^{'} s the idea so u manipulated

the equation with adding a^{2} - a^{2} (line 2) ?

Commented by mrW1 last updated on 26/Sep/17

I just wanted to get the form

(x - p) (x - q) = 0

Certainly one can directly apply the

known formula :

x_{1, 2} = \frac{- 2 a \pm \sqrt{{(2 a)}^{2} - 4 b}}{2} = - a \pm \sqrt{a^{2} - b}

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