Question

write the quadratic polynomial whose zeros are-3 and 4

Answer 1

Step-by-step explanation:

Given -

• Zeroes are -3 and 4

To Find -

• A quadratic polynomial

Now,

As we know that :-

• α + β = -b/a

→ -3 + 4 = -b/a

→ -(1)/1 = -b/a ..... (i)

And

• αβ = c/a

→ -3 × 4 = c/a

→ -12/1 = c/a .... (ii)

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

a = 1

b = -1

c = -12

As we know that :-

For a quadratic polynomial -

• ax² + bx + c

→ (1)x² + (-1)x + (-12)

→ x² - x - 12

Hence,

The quadratic polynomial is x² - x - 12

Answer 2

$\star \; {\underline{\underline{\rm{\blue{Asnwer:}}}}}$

• $\sf{Zeroes\; of\; quadratic \;polynomial\; are\; -3 \;and \;4.}$

$\star \; {\underline{\underline{\rm{\orange{To\;Find:}}}}}$

• $\sf{A \;Quadratic \;Polynomial.}$

$\star \; {\underline{\underline{\rm{\green{Solution:}}}}}$

If $\sf{ \alpha \; and\; \beta \; are \; zeroes \; of \; polynomial \; then,}$

☯ ${\underline{\underline{\boxed{\sf{\purple{\dag \; Sum\;of\;zeroes\; ( \alpha + \beta ) = \dfrac{-b}{a}}}}}}}$

☯ ${\underline{\underline{\boxed{\sf{\purple{\dag \; Product\;of\;zeroes\; ( \alpha \beta ) = \dfrac{c}{a}}}}}}}$

$\rule{200}{2}$

$\small\dag \; {\underline{\sf{\red{Sum\;of\;zeroes:}}}}$

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

$: \implies \sf{-3 + (4) = \dfrac{-b}{a}}$

$: \implies \sf{ \dfrac{-1}{1} = \dfrac{-b}{a} \; \; \; \; ---- eq.(1)}$

$\small\dag \; {\underline{\sf{\red{Product\;of\;zeroes:}}}}$

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

$: \implies \sf{(-3)(4) = \dfrac{c}{a}}$

$: \implies \sf{\dfrac{-12}{1} = \dfrac{c}{a} \; \; \; \; ---- eq.(2)}$

$\rule{200}{2}$

$\small\dag \; {\underline{\sf{\pink{From\;eq(1)\;and\;eq(2):}}}}$

$: \leadsto \sf{a = 1}$

$: \leadsto \sf{b = -1}$

$: \leadsto \sf{c = -12}$

$\rule{200}{2}$

☯ ${\underline{\underline{\boxed{\sf{\purple{\dag \; Quadratic\; polynomial\;:\;ax^2 + bx +c}}}}}}$

$\scriptsize \; \; \; \;\; \; \;\; \; \dag \; {\underline{\sf{\pink{Putting\;value\;of\;a,b\;and\;c\;:}}}}$

$: \implies \sf{(-1)x^2 + (1)x + (-12)}$

$: \implies \sf{- x^2 + x -12}$

$\bold{\underline{\underline{\sf{\red{\dag \; Hence \; Solved!}}}}}$

$\rule{200}{3}$