Question

determine the condition for the roots of the equation ax square + bx + c is equals to zero to differ by 2​

Answer 1

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

Let 'α' and 'β' are two zeroes of equation ax² + bx + c = 0, then

We know that

$\boxed{\purple{\tt Sum\ of\ the\ zeroes= \alpha + \beta = \frac{-b}{a}}}$

OR

$\boxed{\red{\sf Sum\ of\ the\ zeroes= \alpha + \beta = \frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

And

$\boxed{\purple{\tt Product\ of\ the\ zeroes= \alpha \beta = \frac{c}{a}}}$

OR

$\boxed{\red{\sf Product\ of\ the\ zeroes= \alpha \beta = \frac{Constant}{coefficient\ of\ x^{2}}}}$

Now,

According to statement,

we have

$\rm :\implies\: | \alpha - \beta | = 2$

On squaring both sides, we get

$\rm :\implies\: { | \alpha - \beta | }^{2} = {2}^{2}$

$\rm :\implies\: { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta = 4$

$\rm :\implies\: {( \alpha + \beta )}^{2} - 2 \alpha \beta - 2 \alpha \beta = 4$

$\rm :\implies\: { (\alpha + \beta) }^{2} - 4 \alpha \beta = 4$

$\rm :\implies\:\dfrac{ {b}^{2} }{ {a}^{2} } - \dfrac{4c}{a} = 4$

$\rm :\implies\:\dfrac{ {b}^{2} - 4ac }{ {a}^{2} } = 4$

$\rm :\implies\: {b}^{2} - 4ac = 4{a}^{2}$

is the required condition.