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Answer 1

Answer:

answer 1245

Step-by-step explanation:

Answer 2

Answer:

$Answer♡$

Step-by-step explanation:

Answer ::

\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}

↦Firstlylet

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sunderstandtheconceptused

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Here the concept of Quadratic polynomials has been used. If we are given a quadratic polynomial in the form of p(x) = ax² + bx + c then its zeroes will be α and β.

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★ Formula Used :-

$\: \: \large{\boxed{\boxed{\sf{\alpha \: + \: \beta \: = \: \bf{\dfrac{(-b)}{a}}}}}}$

$</p><p>\: \: \large{\boxed{\boxed{\sf{\alpha \: \times \: \beta \: = \: \bf{\dfrac{c}{a}}}}}}$

]

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★ Question :-

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively-1/4,4.

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★ Solution :-

Given,

» Sum of the zeroes, α + β = -(¼)

» Product of zeroes, αβ = 4

• Let the required quadratic polynomial be

p(x) = ax² + bx + c

whose zeroes are α and β.

Here, a is the coefficient of x², b is the coefficient of x and c is the constant term.

Then, according to the question,

$</p><p>~ Case I :-</p><p></p><p>\: \: \large{\sf{\longrightarrow \: \: \: \alpha \: + \: \beta \: = \: \bf{\dfrac{(-b)}{a}}}}⟶α+β= </p><p>a</p><p>(−b)$

$\: \: \large{\sf{\longrightarrow \: \: \: \dfrac{(-1)}{4} \: = \: \bf{\dfrac{(-b)}{a}}}}⟶$

On comparing LHS and RHS, we get,

➣ a = 4 and b = 1 (since, -b = -1)

$\: \: \large{\boxed{\boxed{\tt{a \; = \; 4 \: \: \: and \: \: \: b \; = \; 1}}}}$

$~ Case II :-</p><p></p><p>\: \: \large{\sf{\longrightarrow \: \: \: \alpha \: \times \: \beta \: = \: \bf{\dfrac{c}{a}}}}⟶α×β= </p><p>a</p><p>c$

$\: \: \large{\sf{\longrightarrow \: \: \: \dfrac{4}{1} \: = \: \bf{\dfrac{c}{a}}}}⟶$

Here since, we got a = 4 , earlier, we have to make a here also equal to that.

Then multiplying numerator and denominator by 4, we get,

$\: \: \large{\sf{\longrightarrow \: \: \: \dfrac{4}{1} \: \times \: \dfrac{4}{4} \: = \: \bf{\dfrac{c}{a}}}}⟶$

$\begin{gathered} \: \\ \: \large{\sf{\longrightarrow \: \: \: \dfrac{16}{4} \: = \: \bf{\dfrac{c}{a}}}}\end{gathered} </p><p>⟶$

On comparing, LHS and RHS, we get,

$</p><p>\: \: \large{\boxed{\boxed{\tt{a \; = \; 4 \: \: \: and \: \: \: c \; = \; 16}}}}$

By applying these values in the standard form of quadratic polynomial, we get,

=> p(x) = 4x² + x + 16

$\: \: \large{\underline{\underline{\rm{\leadsto \: \: Thus, \: the \: required \: quadratic \: polynomial \: is \: \boxed{\bf{p(x) \; = \; 4x^{2} \: + \: x \: + \: 16 \: }}}}}}$

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$\: \: \qquad \qquad \large{\underline{Let's \: know \: more \: :-}} </p><p>Let </p><p>′$

• Different types of polynomials are :-

Linear Polynomial

Quadratic Polynomial

Cubic Polynomial

Bi - Quadratic Polynomial