Question

If $x=\frac{2}{3}$ and x = −3 are the roots of the equation ax² + 7x + b = 0, find the values of a and b.

Answer 1

SOLUTION :

Given : ax² + 7x + b = 0 ……………(1)

Here, a and b are unknown constants. Since , x = ⅔ & x = - 3 are the roots of the given equation, so it will satisfy the given equation.

On putting x = 2/3 & x = -3 in eq 1 one by one,

ax² + 7x + b = 0

a(⅔)² + 7(⅔) + b = 0   [x = ⅔]

a(4/9) + 14/3 + b = 0

4a /9 + 14/3 + b = 0

(4a + 14 × 3 + 9 ×b) /9 = 0

(4a + 42 + 9b) = 0 × 9

4a + 42 + 9b = 0  

4a =  - 42 -  9b

a =  (- 42 -  9b)/4 …………..(2)

ax² + 7x + b = 0

a(-3)² + 7(-3) + b = 0    [x = - 3]

9a - 21 + b = 0

9(- 42 -  9b)/4 - 21 + b = 0  

[From eq 2 ]

(-378 - 81b )/4 - 21 + b = 0

(-378 - 81b )  =  4(21 -  b)  

-378 - 81b = 84  -  4b

-81b + 4b = 84 +378

- 77b = 462  

b = 462/-77 = -6  

b = - 6

On substituting b = -6 in eq 2 ,

a =  (- 42 -  9b)/4

a = ( - 42 - 9 × -6)/4

a = (- 42 + 54)/4

a = 12/4 = 3

a = 3  

Hence, the values of a and b are 3 & - 6 .

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\huge{HEY\:THERE}$

since,,,,

x = - 3 and $x=\frac{2}{3}$

then they are given by,,,

(x + 3)($x-\frac{2}{3}$)


(x+3)($x-\frac{2}{3}$)

($x^{2}-\frac{2}{3}$x + 3x - 2 =0..

$3x^{2}$ - 2x + 9x - 6 = 0

$3x^{2}$ + 7x - 6 =0..

comparing with the equation,,.

ax² + 7x + b = 0...

a = 3,, b = -6..

$\bold{REGRDSM\:@\:STYLGG}$