Question

If x=2/3 and x=3 are the roots of the equation ax²+7x+b=0 , find the values of a and b.

Answer 1

Step-by-step explanation:

Given equation is ax^2+7x+b=0 and

roots are x=2/3 and x=3

Now, substitute x=2/3 in the above equation

• a(2/3)^2+7(2/3)+b=0
• (4/9)a+14/3+b=0
• (4/9)a+b=-14/3
• (4a+9b)/9=-14/3
• 4a+9b=-126/3
• 4a+9b=-42

Now, substitute x=3 in the above equation

• a(3)^2+7(3)+b=0
• 9a+21+b=0
• 9a+b=-21

Now, multiply by 9 on both sides,we get

• 81a+9b=-189

Now, substract 81a+9b=-189 from 4a+9b=-42, we get

• -77a=147
• a=-147/77

substitute a=-147/77 in the equation 9a+b=-21,we get

• (-1323/77)+21+b=0
• ((-1323+1617)/77)+b=0
• (294/77)+b=0
• b=-294/77

The values of a and b are -147/77 and -294/77

Answer 2

Given:-

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

To find:-

• Find the values of a and b...?

Solutions:-

• ax² + 7x + b = 0
• x = 2/3 or x = -3

We have to fine a and b.

Now,

If x = 2/3 is a root of the equation.

=> ax² + 7x + b = 0

=> a(2/3)² + 7(2/3) + b = 0

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

=> (4a + 42 + 9b)/9 = 0

=> a = (-9b - 42)/4 ..........(i).

Also, if x = -3 is a root of the equation.

=> ax² + 7x + b = 0

=> a(-3)² + 7(-3) + b = 0

=> 9a - 21 + b = 0 .............(ii).

Now, the multiple equation (ii). by 9 and then Subtract equation (i). we get.

=> 81a + 9b - 189 - 4a - 9b - 42 = 0

=> 77a - 231 = 0

=> a = 231/77

=> a = 3

Now, putting the value of a in Eq. (ii).

=> 9a + b - 21 = 0

=> 9(3) + b - 21 = 0

=> 27 + b - 21 = 0

=> 6 + b = 0

=> b = - 6

Hence, the value of a is 3 and b is -6.