Question

if alpha and beta are the zeros of a quadratic polynomial such that alpha + beta is equals to 24 and Alpha minus beta is equal to 8 find the quadratic polynomial having alpha and beta as its zeros verify the relationship between zeroes and coefficients of the polynomial

Answer 1

Answer:

The required polynomial is $P(x)=x^2-24x+128$.

Step-by-step explanation:

It is given that α and β are the zeros of a quadratic polynomial such that

$\alpha +\beta =24$          .... (1)

$\alpha -\beta =8$             .... (2)

Add both the equations.

$2\alpha=32$

$\alpha=16$

Put this value in equation (1).

$16+\beta =24$

$\beta =8$

The value of α and β are 16 and 8 respectively.

If α and β are the zeros of a quadratic polynomial, the polynomial is in the form of

$P(x)=x^2-(\alpha +\beta)x+\alpha \beta$

$P(x)=x^2-(24)x+16 \times 8$

$P(x)=x^2-24x+128$

Therefore, the required polynomial is $P(x)=x^2-24x+128$.

Answer 2

Step-by-step explanation:

The required polynomial is P(x)=x^2-24x+128P(x)=x

2

−24x+128 .

Step-by-step explanation:

It is given that α and β are the zeros of a quadratic polynomial such that

\alpha +\beta =24α+β=24 .... (1)

\alpha -\beta =8α−β=8 .... (2)

Add both the equations.

2\alpha=322α=32

\alpha=16α=16

Put this value in equation (1).

16+\beta =2416+β=24

\beta =8β=8

The value of α and β are 16 and 8 respectively.

If α and β are the zeros of a quadratic polynomial, the polynomial is in the form of

P(x)=x^2-(\alpha +\beta)x+\alpha \betaP(x)=x

2

−(α+β)x+αβ

P(x)=x^2-(24)x+16 \times 8P(x)=x

2

−(24)x+16×8

P(x)=x^2-24x+128P(x)=x

2

−24x+128

Therefore, the required polynomial is P(x)=x^2-24x+128P(x)=x

2

−24x+128 .