Question

if (x-3) and (x-1/3) are the factors of ax^2 +5x +b then show that a=b​

Answer 1

Answer:

a = b = -3/2

Step-by-step explanation:

ax² + 5x + b

(x - 3) is a factor.

x - 3 = 0

x = 3

Substituting the value of x in the equation ax² + 5x + b = 0

9a + 15 + b = 0 ----------------------> (1)

(x - 3) is a factor.

x - 1/3 = 0

x = 1/3

Substituting the value of x in the equation ax² + 5x + b = 0

a/9 + 5/3 + b = 0

a + 15 + 9b = 0 -----------------------> (2)

Solving 1 and 2

   9a + 15 + b = 0

-9 (a  + 15 + 9b = 0)

_______________

9a + 15 + b - 9a - 135 - 81 b = 0

80 b + 120 = 0

b = -120/80

∴ b = -3/2

∴ a = -3/2

Hope this helps! :)

Answer 2

Answer:

Given:

•  (x-3) and (x-1/3) are the factors of ax^2 +5x +b

To prove:

• a=b​  

Solving Question:

 we are given the factors of the polynomial,thus by substituting the value of 'x' we would reach to our answer.

∴ x - 3 = 0

⇒ x = 3

and '

∵ x - $\dfrac{1}3}=0\\\\\implies x = \dfrac{1}{3}$

Solution:

$substitue\:the\:value\:'x'=3\\\\\implies 9*a+5*3+b=0\\\\\implies 9a+b+15=0\\\\\implies 9a+b = -15........equ(1)$

Then, take x = 1/3

$a(\dfrac{1}{3})^2+5*\dfrac{1}{3}+b=0\\\\\implies \dfrac{a}{9}+\dfrac{5}{3}+b=0\\\\\implies \dfrac{a}{9}+b=\dfrac{5}{3}(\:multiply\:9)\\\\\implies a+9b=-15 .......equ(2)$

Note; when we substitute the value of 'x it will become 0 because it is a factor.

Take equ(1) and(2)

the R.H.S of the both are equal. therefore

⇒ 9a + b = a+ 9b

⇒ 8a = 8b

⇒ a= b

Hence proved