Question

In a quadratic equation with leading coefficient 1 , a student reads the coefficient 16 of x wrongly as 19 and

obtain the roots as -15 and -4 .The correct roots are
:-​

Answer 1

Step-by-step explanation:

$\sf Since\:the\:root\: obtained\:are\: -15 \: and \: - 4 \\ \sf the \:equation\:will\:be \\ \\ \sf ( x + 15 )( x + 4 ) = 0 \\ \\ \sf \therefore {x}^{2} + 19x + 60 = 0 \\ \\ \sf Now\:putting\:the\:correct\: coefficient\:of \: x \: we \: get \\ \\ {x}^{2} + 16x + 60 = 0 \\ \\ \sf \therefore {x}^{2} + 10x + 6x + 60 = 0 \\ \\ \sf \therefore ( x + 10 )( x + 6 ) = 0 \\ \\ \sf \therefore The\: correct\:roots \:are \: - 10 \: and \: - 6$

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