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Answer 1

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Answer 2

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Given, x = 2/3 and x = – 3 are the roots of the quadratic equation ax^2 + 7x + b = 0.

${ax}^{2} + 7x + b = 0 \: \: \: \: - - - - - - (1)$


Putting x= 2/3 in equation (1), we get

$a( { \frac{2}{3} ) }^{2} + 7( \frac{2}{3} ) + b = 0 \\ \\ = > \frac{4}{9} a + \frac{14}{3} + b = 0 \\ \\ = > 4a + 42 + 9b = 0 \\ \\ = > 4a + 9b = - 42 \: \: \: \: \: - - - - (2)$


Putting x = – 3 in equation (1), we get

$a( { - 3) }^{2} + 7( - 3) + b = 0$
⇒ 9a – 21 + b = 0

⇒ 9a + b = 21

⇒ b = 21 – 9a ...(3)

Solving (2) and (3), we get

4a + 9 ( 21 – 9a) = – 42

⇒ 4a + 189 – 81a = – 42

⇒ –77a = – 42 – 189 = – 231

⇒ a = 3

Putting the value of a in (3), we get

b = 21 – 9 × 3 = 21 – 27 = – 6

Thus, the value of a and b is 3 and – 6 respectively.


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# Nikky