Question

The sum of the roots of the equation mx2-(m+n)x+1=0

Answer 1

$\large \bold \red{ \underline{ \underline{answer}}} \\ sum \: \: of \: \: roots \: = \frac{m + n}{m}$

$\large \bold \green{ \underline{ \underline{question}}} \\ the \: sum \: \: of \: \: the \: \: roots \: \: of \: \: the \: \: equation \: \: \\ m {x}^{2} - (m + n)x + 1 = 0 \\ \large \bold \blue{ \underline{ \underline{step \: . \: by \: . \: step \: . \: explanation}}} \\ sum \: of \: roots = \frac{ - b}{a} = \frac{m + n}{m} \\ \large \bold \green{notes = } \\ products \: of \: roots = \frac{c}{a} = \frac{1}{m} \\ \\ \large \bold \pink{formula \: of \: \: quadratic \: \: polynomial} \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta$

Answer 2

QUESTION:

The sum of the roots of the equation mx2-(m+n)x+1=0

CONCEPT:

We know that; quadratic polynomial is in the form of

$a {x}^{2} + bx + c = 0$

where;

b = sum of zeroes

c = product of zeroes

in other form we can write the equation;

${x}^{2} - (sum \: of \: zeroes)x + product \: of \: zeroes)$

$(remember \: before \: sum \: of \: zeroes \: there \: is \: always \: negative \: sign)$

We known that;

$sum \: of \: zeroes = \frac{ - coefficient \:of \: x }{coeffiecient \: of \: {x}^{2} } = \frac{ - b}{a}$

now come to main question;

$m {x}^{2} - (m + n)x + 1 = 0$

so,

here;

b = -m-n

c = m

using the formula;

$</u><u>\</u><u>sum</u><u>\: of \: zeroes = \frac{ - ( - m - n)}{m} \\ sum \: of \: zeroes = \frac{m + n}{m} </u><u>$

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