Question

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Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: given \: quadratic \: equation \: is \: 20ax \: square(2a - b square\:)x - b \: square = 0$

$so \: 2ax \: square - 2ax - bsquare \: x - b \: square = 0$

$2ax(x + 1)-bsquare(x + 1) = 0$

$ie \: (2ax - b \: square)(x + 1) = 0 \\ 2ax - b \: square = 0 \: or \: x + 1 = 0 \\ ie \: 2ax =b \: square \: or \: x = - 1 \\ so \: thus \\ \\ x = b \: square \div 2a \: or \: x = - 1$

$hence \: bsquare \div 2a \: and \: - 1 \: are \: roots \: of \: above \: quadratic \: equation$