Question

5. If x = 2/3 and x = -3 are the roots of the equation ax? + 7x + b = 0, find the values of a
and b.
​

Answer 1

EXPLANATION.

x = 2/3  and  x = -3 are the roots of the equation,

⇒ ax² + 7x + b = 0.

As we know that,

Put the value of x = 2/3 in equation, we get.

⇒ a(2/3)² + 7(2/3) + b = 0.

⇒ a(4/9) + 14/3 + b = 0.

⇒ 4a/9 + 14/3 + b = 0.

Taking L.C.M in equation, we get.

⇒ 4a + 42 + 9b = 0. ⇒ (1).

Put the value of x = -3 in equation, we get.

⇒ a(-3)² + 7(-3) + b = 0.

⇒ a(9) - 21 + b = 0.

⇒ 9a - 21 + b = 0.

⇒ b = 21 - 9a ⇒ (2).

Put the value of equation (2) in equation (1), we get.

⇒ 4a + 42 + 9(21 - 9a) = 0.

⇒ 4a + 42 + 189 - 81a = 0.

⇒ 231 - 77a = 0.

⇒ 77a = 231.

⇒ a = 3.

Put the value of a = 3 in equation (2), we get.

⇒ b = 21 - 9(3).

⇒ b = 21 - 27.

⇒ b = -6.

Values of A = 3 & B = -6.

Answer 2

$\\ \: \sf \large \color{navy}\underline{ Given :-}$

⇒  x = 2/3 and x = -3 are the roots of the equation ax² + 7x + b = 0

$\\ \: \sf \large \color{navy}\underline{ \:To \: Find :-}$

⇒ Values of a and b

$\\ \: \sf \large \color{navy}\underline{ \: Required \: Solution :-}$

→ Firstly we will put the value of x = 2/3 in the given equation .

$\sf \implies a(\dfrac{2}{3})^{2} + 7(\dfrac{2}{3}) + b = 0$

$\sf \implies a \times \dfrac{4}{9} + \dfrac{14}{3} +b = 0$

$\sf \implies \dfrac{4a}{9} + \dfrac{14}{3} +b = 0$

$\sf \implies 4a+42 + 9b = 0 \; -- (i)$

→ Now put the value of x = -3 in the given equation

$\sf \implies a(-3)^{2} + 7(-3) + b = 0$

$\sf \implies a \times 9 - 21 + b = 0$

$\sf \implies 9a - 21 + b = 0$

$\sf \implies b = 21 - 9a \; --- (ii)$

___________

→ Let's put the value of eq ( ii ) in eq ( i )

$\sf \implies 4a + 42 + 9(21 - 9a) = 0$

$\sf \implies 4a + 42 + 189 - 81a = 0$

$\sf \implies 231 - 77a = 0$

$\sf \implies 77a = 231$

$\sf \implies a = \dfrac{231}{77} = 3$

___________

→ Let's put the value of a = 3 in eq ( ii )

$\sf \implies b = 21 - 9 \times 3$

$\sf \implies b = 21 - 27$

$\sf \implies b = -6$

_________

∴ Here ,  a = 3  , b = -6