Question

★ Question:-

If the product of the zeros of quadratic polynomial p(x) is - 81 and one of the zeros is 3 then p(x) = _________

(a) x² + 24x - 81

(b) x² - 24x - 81

(c) x² - 24x + 81

(d) x² + 24x + 81

↦Explanation needed!! ​

Answer 1

Answer:

$\alpha \beta = - 81$

$let \: \alpha = 3$

Putting the value of alpha in the product :

$\alpha \beta = - 81$

$3 \beta = - 81$

$\beta = - 81 \div 3$

$\beta = - 27$

Now,

$\alpha + \beta = 3 + ( - 27)$

$\: \: 3 - 27$

$\: \: - 24$

$p(x) = {x}^{2} - ( \alpha + \beta )x + \alpha \beta$

$\: \: {x}^{2} - ( - 24)x + ( - 81)$

$\: \: {x}^{2} + 24x - 81$

Ans. a) x^2 + 24x - 81

Hope it helps.

Answer 2

Step-by-step explanation:

Answer:

\alpha \beta = - 81αβ=−81

let \: \alpha = 3letα=3

Putting the value of alpha in the product :

\alpha \beta = - 81αβ=−81

3 \beta = - 813β=−81

\beta = - 81 \div 3β=−81÷3

\beta = - 27β=−27

Now,

\alpha + \beta = 3 + ( - 27)α+β=3+(−27)

\: \: 3 - 273−27

\: \: - 24−24

p(x) = {x}^{2} - ( \alpha + \beta )x + \alpha \betap(x)=x

2

−(α+β)x+αβ

\: \: {x}^{2} - ( - 24)x + ( - 81)x

2

−(−24)x+(−81)

\: \: {x}^{2} + 24x - 81x

2

+24x−81

Ans. a) x^2 + 24x - 81

Hope it helps.