Question

Find the zeroes of the quadratic polynomial abx2+(b2-ac)x-bc, and verify the relationship between zeroes and its coefficients.

Answer 1

hope this helps you ☺☺

Answer 2

$Let \: p(x) = abx^{2} + (b^{2}-ac)x - bc$

/* By splitting the middle term we see */

$p(x) = abx^{2} + b^{2} x -acx - bc\\= bx( ax + b) -c(ax + b) \\= (ax + b)( bx - c)$

$p(x) = abx^{2} + (b^{2}-ac)x - bc \:is \:zero \\when\: either \: ax + b = 0 \: Or \: bx - c = 0$

$i.e., when \: x = \frac{-b}{a } \: Or \: x = \frac{c}{b}$

$The \: zeroes \: of \: abx^{2} + (b^{2}-ac)x - bc \\are \: \frac{-b}{a } \:and \: \frac{c}{b}$

$We \:can \: see \: that \: the :$

$Sum \:of \:the \: zeroes \\= \frac{-b}{a } + \frac{c}{b}\\= \frac{-b^{2} + ac}{ab} \\= \frac{-(b^{2} -ac)}{ab} \\= \frac{ - ( Coefficient \:of \: x )}{ Coefficient \:of \:x^{2}}$

$Product\:of \:the \: zeroes \\= \frac{-b}{a } \times \frac{c}{b}\\= \frac{-c}{a} \\= \frac{ Constant \:term}{ Coefficient \:of \:x^{2}}$

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