Question

Question:-

Given roots of equation:-

`\sf x = (2)/(3) \: nd \: x=-3`
Find a nd b in ax²+7x+b =0



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Need lil help ~
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Answer 1

`\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}`

given - an equation and its roots !

To find - values of a and b in the given equation ~

`Quadratic \: eq {}^(n) = ax {}^(2) + 7x + b = 0 \\`

roots of the question are
`\dashrightarrow - 3 \: and \: (2)/(3) \\`

Now ,

`Sum \: of \: zeroes \: ( \: \alpha \: + \: \beta \: ) = - ( - b)/(a) \\ \\ \implies \: - 3 + (2)/(3) = ( - 7)/(a) \\ \\ \implies \: ( - 9 + 2)/(3) \\ \\ \implies ( - 7)/(3) = ( - 7)/(a) \\ \\ \underline{on \: comparing} \\ \\\bold\red{\implies \:a = 3}`

further ,

`Product \: of \: zeroes \: ( \: \alpha \: * \beta \: ) = (c)/(a) \\ \\ \implies \: (- \cancel {3})( (2)/(\cancel3) ) = (b)/(a) \\ \\ \implies \: - 2 = (b)/(3) \\ \\ \implies \: b = 3 * - 2 \\ \\ \bold\red{\implies \: b = - 6}`

hope helpful :D