Question

If a and ß are zeroes of the polynomial x2 - 6x + a. Find the value of a if 3a + 2ß = 20​

Answer 1

Following is the value of a 3alpha + 2beta = 20

Given

${x}^{2} - 6x + a = 0$

α,ß are the zeroes of the polynomial

=> α+ß=6-----(i)

αß=a------(ii)

Given

3α+2ß=20------(iii)

Solving (i) and (iii) we get

α=8

ß=-2

Substituting these values in (ii) we get

=> (8)(-2)=a

=> a=-16

Answer 2

Step-by-step explanation:

Given :-

α and ß are zeroes of the polynomial x² - 6x + a. and 3α + 2ß = 20

To find :-

Find the value of a ?

Solution :-

Given Quadratic Polynomial is x²-6x+a

Let P(x) = x²-6x+a

On Comparing this with the standard quadratic Polynomial ax²+bx+c

a = 1

b = -6

c = a

Given zeores are α and ß

We know that

The sum of the zeroes = -b/a

=>α+ß = -(-6)/1

=> α+ß = 6

=>α = 6-ß -------------(1)

The product of the zeroes = c/a

=> αß = a/1

=> αß = a --------------(2)

Given that

3α+2ß = 20

From (1)

=> 3(6-ß)+2ß = 20

=> 18 -3ß +2ß = 20

=> 18-ß = 20

=> ß = 18-20

=> ß = -2

On Substituting the value of ß in (1) then

α = 6-(-2)

=> α = 6+2 = 8

Now,

From (2)

=>a = (8)(-2)

=> a = -16

Answer:-

The value of a for the given problem is -16

Used formulae:-

→ The standard quadratic Polynomial ax²+bx+c

→ The sum of the zeroes = -b/a

→ The product of the zeroes = c/a