Question

$\huge\color{teal}{\bold{\underline{\underline{Question:-}}}}$

If α , β are the roots of ax² + 2bx + c = 0 and α + δ , β + δ are the roots of Ax² + 2Bx + C = 0 , then
$\large\rm{\frac{b^2-ac}{B^2-AC}}$ is

1] $\large\rm{\frac{a}{A}}$

2] $\large\rm{\frac{a^2}{A^2}}$

3] $\large\rm{\frac{a^3}{A^3}}$

4] None of these ​

Answer 1

question:

•alpha , beta are the roots of ax²+2bx+c=0

•alpha+Delta, beta+Delta are the roots of Ax²+2Bx+C=0

to find:

the ratio of $\tt \dfrac{b^2-ac}{B^2-AC}$

solution:

since we see alpha as well as beta common in both equations,

we will try to convert this equation into a equation wherein the roots appear similar.

$\boxed{\red{\tt difference of roots = ✓4b^2-4ac/a^2}}$

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you can derive this by squaring alpha-beta as:

$\tt (alpha-beta)²=alpha²+beta²-2alphabeta$

$\tt =(-b/a)²-4c/a$

$\tt =√4b²-4ac/a²$

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so,

let's take the ratio of both equations as difference of roots

$\tt alpha-beta=√4b²-4ac/a²$

$\tt alpha+delta-beta-delta=√4B²-4AC/A²$

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taking ratio,

$\tt \dfrac{alpha-beta}{alpha-beta} =\dfrac{ 4b²-4ac×A²}{a²×4B²-4AC}$

$\tt 1 =\dfrac{4b²-4ac×A²}{a²×4B-4AC}$

$\tt \dfrac{a²}{A² }=\dfrac{ 4b²-4ac}{4B-4AC}$

$\boxed{\tt \dfrac{a²}{A²}= \dfrac{4(b²-4ac)}{4(B-AC)}}$

cancel the 4.

so, option 2 that is a²/A² is correct

Answer 2

Questions:-

If α , β are the roots of ax² + 2bx + c = 0 and α + δ , β + δ are the roots of Ax² + 2Bx + C = 0 , then

$\large\rm{\frac{b^2-ac}{B^2-AC}} is$

Solution:-

$\rm \: a {x}^{2} + 2bx + c = 0 \: having \: roots \: \alpha \: and \: \beta$

$\rm \: \alpha + \beta = - \dfrac{2b}{a} \: \alpha \beta = \dfrac{c}{a}$

We know

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

put the values, we get

$\rm \implies \: { (\alpha - \beta )}^{2} = - {( \dfrac{2b}{a} )}^{2} - 4 \times \dfrac{c}{a}$

$\implies \rm { (\alpha - \beta )}^{2} = \dfrac{ {4b}^{2} }{ {a}^{2} } - \dfrac{4c}{a}$

$\rm =\dfrac{4}{a}(\dfrac{ {b}^{2} }{a} - c) =\dfrac{4}{a} \dfrac{ {b}^{2} - ac }{a} =\dfrac{4}{ {a}^{2} }({b}^{2}- ac)$

α + δ , β + δ are the roots of Ax² + 2Bx + C = 0

$\rm\alpha + \delta \: + \beta + \delta = \dfrac{2B}{ A}$

$\rm( \alpha + \delta)( \beta + \delta) = \dfrac{C}{A}$

$\rightarrow\rm { [(\alpha + \delta) - ( \beta + \delta) ] }^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \\ = \rm { [(\alpha + \delta) + ( \beta + \delta) ] }^{2} - 4(\alpha + \delta) + ( \beta + \delta)$

$\rm { (\alpha - \beta )}^{2} = { (\dfrac{2B}{A}) }^{2} - 4 \dfrac{C}{A}$

$\rm= \dfrac{4 {B}^{2} }{ {A}^{2} } - \dfrac{4C}{A} = \dfrac{4}{ {A}^{2} }( {B}^{2} - AC)$

$\rightarrow\rm \frac{\dfrac{4}{ {a}^{2} }({b}^{2}- ac)}{ \dfrac{4}{ {A}^{2} }( {B}^{2} - AC)} = 1$

$\implies\rm \ \: \dfrac{ \cancel4}{ {a}^{2} }({b}^{2}- ac) \times \dfrac{ {A}^{2} }{ \cancel4 ({B}^{2} -AC )} = 1$

$\implies \rm\dfrac{ {b}^{2} - ac}{ {B}^{2} -AC } = \dfrac{ {a}^{2} }{ {A}^{2} } = { \bigg(\dfrac{a}{A} \bigg )}^{2}$

$2] \large\rm{\dfrac{a^2}{A^2}} \: is \: the \: correct \: answer.$