Question

Plz answer this question with step by step explanation.​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

if the ratio of roots of the equation x² + px + q = 0 be equal to the ratio of the equation x² + bx + c = 0 , then Find the value of (p²c - b²q) = ?

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\green{Given}}}}}$

• equation x² + px + q = 0 ...(1)
• equation x² + bx + c = 0 ....(2)

$\Large{\underline{\mathfrak{\bf{\green{Condition}}}}}$

• Ratio of the roots of the both equation are equal .

$\large{\underline{\mathfrak{\bf{\orange{Step-by-Step-Explanation}}}}}$

• Root of first equation be α and β
• Root of first equation be α and βRoot of second equation be r and s

For equation first

$\underline{\sf{\orange{\:Using\:Dharacharya\:Formula}}}$

$\small\boxed{\sf{\green{\:x\:=\:\dfrac{-b\:\pm \sqrt{(b^2-4ac)}}{2a}}}}$

where,

• a = 1.
• b = p.
• c = q.

Keep value ,

$\mapsto\sf{\:x\:=\:\dfrac{-p\:\pm \sqrt{(p^2-4.1.q)}}{2.1}} \\ \\ \mapsto\sf{\:x\:=\:\dfrac{-p\:\pm \sqrt{(p^2-4q)}}{2}}$

First Take, ( + ve ) Sign,

First Factor will be ,

$\mapsto\sf{\orange{\:\alpha\:=\:\dfrac{-p\:+\:\sqrt{(p^2-4q)}}{2}}}$

Now, Take ( - ve ) Sign,

Second Factor will be ,

$\mapsto\sf{\orange{\:\beta\:=\:\dfrac{-p\:-\:\sqrt{(p^2-4q)}}{2}}}$

_____________________

For Equation Second

$\small\boxed{\sf{\green{\:x\:=\:\dfrac{-B\:\pm \sqrt{(B^2-4AC)}}{2a}}}}$

where,

• A = 1.
• B = b
• C = c.

Keep value,

$\mapsto\sf{\:x\:=\:\dfrac{-b\pm \sqrt{(b^2-4c)}}{2}}$

First, Take ( + ve ) Sign,

First Factor will be

$\mapsto\sf{\:r\:=\:\dfrac{-b\:+\:\sqrt{(b^2-4c)}}{2}}$

Now, Take ( - ve ) Sign,

Second Factor will be

$\mapsto\sf{\:s\:=\:\dfrac{-b\:-\:\sqrt{(b^2-4c)}}{2}}$

_____________________

A/c to question,

( Ratio of the roots of the both equation are equal )

$\small\boxed{\sf{\green{\:\alpha:\beta\:=\:r:s\:=\:\dfrac{\alpha}{\beta}\:=\:\dfrac{r}{s}}}} \\ \\ \mapsto\sf{\:\dfrac{\dfrac{-p\:+\:\sqrt{(p^2-4q)}}{2}}{\dfrac{-p\:-\:\sqrt{(p^2-4q)}}{2}}\:=\:\dfrac{\dfrac{-b\:+\:\sqrt{(b^2-4c)}}{2}}{\dfrac{-b\:-\:\sqrt{(b^2-4c)}}{2}}} \\ \\ \mapsto\sf{\:\dfrac{-p\:+\:\sqrt{(p^2-4q)}}{-p\:-\:\sqrt{(p^2-4q}}\:=\:\dfrac{-b\:+\:\sqrt{(b^2-4c}}{-b\:-\:\sqrt{(b^2-4c)}}}</p><p>$

squaring both side,

$\mapsto\sf{\:\left(\dfrac{-p\:+\:\sqrt{(p^2-4q)}}{-p\:-\:\sqrt{(p^2-4q}}\right)^2\:=\:\left(\dfrac{-b\:+\:\sqrt{(b^2-4c}}{-b\:-\:\sqrt{(b^2-4c)}}\right)^2} \\ \\ \mapsto\sf{\:\dfrac{p^2}{(2q+p\sqrt{p^2-4q})}\:=\:\dfrac{b^2}{(2c+b\sqrt{b^2-4c})}} \\ \\ \mapsto\sf{\red{\:(p^2c-b^2q)\:=\:\dfrac{(b^2p\sqrt{p^2-4q}-p^2b\sqrt{b^2-4c})}{2}}}$

Thus:-

Answer will be Options number (4).

Answer 2

Answer:

Option D pls mark as branliest