Question

Find the quadratic polynomial in each case, with the given numbers as the sum and
product of its zeroes respectively.
()
-1
(1) 12,
(1) O.




(iv) 1,1
(v)-1/4,1/4
(vi) 4, 1​

Answer 1

Step-by-step explanation:

4. o

$4. \\ sum \: of \: zeroes = \alpha + \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 + 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 \\ product \: of \: zeroes \: = \alpha \beta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 \times 1 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1$

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

As, we have to find the quadratic polynomial (Q.P)

↪(a)

We know that,

$\small{\star{\boxed{\sf{Quadratic \: polynomial = x^2 - (Sum)x + Product}}}}$

★ Putting Values ★

$\begin{lgathered}\sf{\rightarrow Q.P = x^2 -(1)x + 1} \\ \\ \sf{\rightarrow Q.P = x^2 - x + 1} \\ \\ \Large{\implies{\boxed{\boxed{\sf{x^2 - x + 1}}}}}\end{lgathered}$

$\rule{200}{2}$

↪(b)

We know that,

$\small{\star{\boxed{\sf{Quadratic \: polynomial = x^2 - (Sum)x + Product}}}}$

★ Putting Values ★

$\begin{lgathered}\sf{\rightarrow Q.P = x^2 - \bigg(\frac{-1}{4}\bigg)x + \frac{1}{4}} \\ \\ \sf{\rightarrow Q.P = x^2 + \frac{1}{4}x + \frac{1}{4}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{x^2 + \frac{1}{4}x + \frac{1}{4}}}}}}\end{lgathered}$

$\rule{200}{2}$

↪(c)

We know that,

$</p><p>\small{\star{\boxed{\sf{Quadratic \: polynomial = x^2 - (Sum)x + Product}}}}$

★ Putting Values ★

$\begin{lgathered}\sf{\rightarrow Q.P = x^2 - (4)x + 1} \\ \\ \sf{\rightarrow Q.P = x^2 - 4x + 1} \\ \\ \Large{\implies{\boxed{\boxed{\sf{x^2 - 4x + 1}}}}}\end{lgathered}$