Question

find a quadratic polynomial whose zeros are 1 and -3 verify the relationship between the coefficients and zeros of the polynomial ​

Answer 1

Step-by-step explanation:

Given -

• Zeroes are 1 and -3

To Find -

• A quadratic polynomial and verify the relationship between the zeroes and the coefficient

Now,

As we know that :-

• α + β = -b/a

→ -3 + 1 = -b/a

→ -2/1 = -b/a ....... (i)

And

• αβ = c/a

→ -3 × 1 = c/a

→ -3/1 = c/a ........ (ii)

Now, From (i) and (ii), we get :-

a = 1

b = 2

c = -3

Now,

As we know that :-

For a quadratic polynomial :-

• ax² + bx + c

→ (1)x² + (2)x + (-3)

→ x² + 2x - 3

Hence,

The quadratic polynomial is x² + 2x - 3

Verification :-

• α + β = -b/a

→ -3 + 1 = -(2)/1

→ -2 = -2

LHS = RHS

And

• αβ = c/a

→ -3 × 1 = -3/1

→ -3 = -3

LHS = RHS

Hence,

Verified...

It shows that our answer is absolutely correct.

Answer 2

Given:-

• $\sf{ \alpha = 1}$
• $\sf{\beta = -3}$

$\bold{\fbox{\color{Purple}{Standard \ form \ of \ Quadratic \ polynomial \ :}}}$

$\implies \sf{x^2 - (sum \; of \; zeros)x + (Product \; of \; zeros)}$

$\implies \sf{x^2 - ( \alpha + \beta )x + (\alpha \beta)}$

$\implies \sf{ x^2 - [ 1 + (-3)]x + [1 × (-3)]}$

$\implies \sf{x^2 - [-2]x + [-3]}$

$\implies \sf{x^2 + 2x -3}$

$\sf \boxed{\dag \; Here \; a = 1 ; \; b = 2 ; \; c = -3}$

$\rule{200}{2}$

★ $\sf{Relationship \; b/w \; zeroes \; and \; coefficient}$

$\bullet \; \sf{sum \; of \; zeros \; :}$

$\implies \sf{ \alpha + \beta = \dfrac{-b}{a}}$

$\implies \sf{ 1 + (-3) = \dfrac{ -2}{1}}$

$\large\bold{\underline{\underline{\boxed{\sf{\pink{-2 = -2}}}}}}$

$\bullet \; \sf{Product \; of \; zeros \; :}$

$\implies \sf{ \alpha × \beta = \dfrac{c}{a}}$

$\implies \sf{ 1 × (-3) = \dfrac{ -3}{1}}$

$\large\bold{\underline{\underline{\boxed{\sf{\pink{-3 = -3}}}}}}$

$\rule{200}{2}$