Question

Fint the zeros of the quadratic polynomial f(x)=abx^2+(b^2-ac)x

Answer 1

Hello dear user!! ✌️✌️


Here is your answer goes like this:


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Your question :


Fint the zeros of the quadratic polynomial f(x)=abx^2+(b^2-ac)x



Answer:


Given;

=>f( x ) = abx² + ( b² - ac )x - bc

To find the zeroes of f( x ) ,lets

take f( x ) = 0, we get,

=>abx² + ( b² - ac )x - bc = 0

=>abx² + b² x - acx - bc = 0

=>bx( ax + b ) - c( ax + b ) = 0

=>( ax + b ) ( bx - c ) = 0

=>ax + b = 0 or bx - c = 0

=>ax = - b or bx = c

=>x = -b/a or x = c/b

Therefore ,and hence

=>-b/a , c/b are two zeroes of f( x ).


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Answer 2

${\Huge{\mathfrak{\purple{Answer:}}}}$

$\begin{lgathered}x = \frac{ -B \pm \sqrt{ {B}^{2} - 4 AC} }{2 A} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) \pm \sqrt{( {b}^{2} - ac {)}^{2} } - 4(ab) ( - bc) }{2ab} \\ \\ \implies \: x \: = \frac{ - ( {b}^{2} - ac) \pm \sqrt{( {b}^{2} - ac {) }^{2} +4a {b}^{2}c } }{2ab} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) \pm \sqrt{ {b}^{4} - 2a {b}^{2}c + {a}^{2} {c}^{2} + 4a {b}^{2}c } }{2ab} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) \pm \sqrt{( {b}^{2} + ac {)}^{2} } }{2ab} \: \: \: \implies \: x = \frac{ - ( {b}^{2} - ac) \pm( {b}^{2} + ac) }{2ab} \\ \\ \implies \: x = \frac{ - ( {b}^{2} - ac) + ( {b}^{2} + ac) }{2ab} \: \: \: \: or \: \: \: \: x = \frac{ - ( {b}^{2} -ac) - ( {b}^{2} + ac) }{2ab} \\ \\ \implies \: x = \frac{2ac}{2ab} \: \: \: \: \: \: or \: \: \: \: x = \frac{ - 2 {b}^{2} }{2ab} \: \: \: \: \: \implies \: x = \frac{c}{b} \: \: \: \: \: \: or \frac{ - b}{a}\end{lgathered} </p><p>$