Question

if alpha and bita are the zero of the quadratic polynomial xquare-6a+a.find the value of a if 3alpha+2beta=20​

Answer 1

Answer:-

Given:-

α & β are the roots of x² - 6ax + a = 0

On comparing the given polynomial with the standard form of a quadratic equation i.e., ax² + bx + c = 0 ;

Let,

• a = 1
• b = - 6a
• c = a.

We know that,

Sum of the roots = - b/a

⟹ α + β = - ( - 6a)/a

⟹ α + β = 6

⟹ α = 6 - β -- equation (1)

It is also given that,

⟹ 3α + 2β = 20

Substituting the value of α from equation (1) we get,

⟹ 3(6 - β) + 2β = 20

⟹ 18 - 3β + 2β = 20

⟹ 18 - 20 = β

⟹ - 2 = β

Hence, β = - 2 is one root of x² - 6ax + a = 0.

So substitute x = - 2 in the given equation.

⟹ ( - 2)² - 6(- 2) * a + a = 0

⟹ 4 + 12a + a = 0

⟹ 4 + 13a = 0

⟹ 13a = - 4

⟹ a = - 4/13

Answer 2

Given :-

α and β are zeroes of x² - 6a + a

To Find :-

Value of a

Solution :-

As we know that

Sum of zeroes = -b/a

Here

b = -6

a = 1

Sum = -(-6)/1

Sum = 6/1

Sum = 6

α + β = 6 (1)

3α + 2β = 20 (2)

On multiplying 1 with 2

2(α + β) = 2(6)

2(α + β) = 12

2α + 2β = 12

Subtracting both

3α + 2β - 2α - 2β = 20 - 12

(3α - 2α) + (2β - 2β) = 8

3α - 2α = 8

α = 8

Using 1

8 + β = 6

β = 6 - 8

β = -2

Now

Product of zeroes = c/a

Here,

c = a

a = 1

αβ = a/1

8 × (-2) = a/1

8 × -2/1 = a

-16/1 = a

-16 = a

Hence

Value of a is -16