Question

finda quadratic polynomial whose zeros are alpha and beta satisfying the relationship alpha + beta = 3 and alpha - beta = -1

Answer 1

Required Quadratic equation:-

x² -3x +2 =0

Given :-

• α + β = 3
• α - β = -1

To find :-

The Quadratic equation which satisfies α + β = 3

α - β = -1

SOLUTION:-

Firstly we find the α , β values from given equation . Then we find the Quadratic equation whose roots are α , β

For finding the α , β values We shall add the both equations

α + β = 3 ---- eq 1

α - β = -1 ------ eq 2

eq 1 + eq 2

α + β + α - β = 3- 1

2α = 2

α = 2/2

α = 1

eq1 - eq 2

α + β -[α -β ] = 3-(-1)

α +β -α +β = 3+1

2β =4

β=4/2

β =2

So, the value of α , β are 1, 2

Now , we shall find the Quadratic equation whose roots are α , β that is

x²-(α + β ) x + α β

Substituting the values ,

x²-(1+2)x + (1)(2)

x² -(3)x + 2

x² -3x + 2

So, the required Quadratic polynomial is x²-3x + 2

Verification:-

As we got the Quadratic equation x²-3x +2 Now , we shall find roots (α ,β ) Since , It should satisfies the conditions

• α+ β = 3
• α -β = -1

x² -3x + 2 = 0

Splitting the middle term,

x² -x -2x +2 = 0

x(x-1) -2(x-1) =0

(x-1)(x-2) = 0

x = 1, 2

Since,

α= 1

β= 2

Verifying the Given conditions :-

α+ β = 3

1 +2 =3

3=3 (Verified)

α -β = -1

1 -2 = -1

-1 = -1 (Verified)

Answer 2

Required Quadratic equation:-

x² -3x +2 =0

Given :-

α + β = 3

α - β = -1

To find :-

The Quadratic equation which satisfies α + β = 3

α - β = -1

SOLUTION:-

Firstly we find the α , β values from given equation . Then we find the Quadratic equation whose roots are α , β

For finding the α , β values We shall add the both equations

α + β = 3 ---- eq 1

α - β = -1 ------ eq 2

eq 1 + eq 2

α + β + α - β = 3- 1

2α = 2

α = 2/2

α = 1

eq1 - eq 2

α + β -[α -β ] = 3-(-1)

α +β -α +β = 3+1

2β =4

β=4/2

β =2

So, the value of α , β are 1, 2

Now , we shall find the Quadratic equation whose roots are α , β that is

x²-(α + β ) x + α β

Substituting the values ,

x²-(1+2)x + (1)(2)

x² -(3)x + 2

x² -3x + 2

So, the required Quadratic polynomial is x²-3x + 2

Verification:-

As we got the Quadratic equation x²-3x +2 Now , we shall find roots (α ,β ) Since , It should satisfies the conditions

α+ β = 3

α -β = -1

x² -3x + 2 = 0

Splitting the middle term,

x² -x -2x +2 = 0

x(x-1) -2(x-1) =0

(x-1)(x-2) = 0

x = 1, 2

Since,

α= 1

β= 2

Verifying the Given conditions :-

α+ β = 3

1 +2 =3

3=3 (Verified)

α -β = -1

1 -2 = -1

-1 = -1 (Verified)