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Answer:
answer 1245
Step-by-step explanation:
Answer:
[tex]Answer♡[/tex]
Step-by-step explanation:
Answer ::
\: \: \boxed{\boxed{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}
↦Firstlylet
′
sunderstandtheconceptused
[tex]
[/tex]
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 :-
[tex]\: \: \large{\boxed{\boxed{\sf{\alpha \: + \: \beta \: = \: \bf{\dfrac{(-b)}{a}}}}}}[/tex]
[tex]
\: \: \large{\boxed{\boxed{\sf{\alpha \: \times \: \beta \: = \: \bf{\dfrac{c}{a}}}}}}[/tex]
]
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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,
[tex]
~ Case I :-
\: \: \large{\sf{\longrightarrow \: \: \: \alpha \: + \: \beta \: = \: \bf{\dfrac{(-b)}{a}}}}⟶α+β=
a
(−b)[/tex]
[tex]\: \: \large{\sf{\longrightarrow \: \: \: \dfrac{(-1)}{4} \: = \: \bf{\dfrac{(-b)}{a}}}}⟶ [/tex]
On comparing LHS and RHS, we get,
➣ a = 4 and b = 1 (since, -b = -1)
[tex]\: \: \large{\boxed{\boxed{\tt{a \; = \; 4 \: \: \: and \: \: \: b \; = \; 1}}}} [/tex]
[tex]~ Case II :-
\: \: \large{\sf{\longrightarrow \: \: \: \alpha \: \times \: \beta \: = \: \bf{\dfrac{c}{a}}}}⟶α×β=
a
c[/tex]
[tex]\: \: \large{\sf{\longrightarrow \: \: \: \dfrac{4}{1} \: = \: \bf{\dfrac{c}{a}}}}⟶ [/tex]
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,
[tex]\: \: \large{\sf{\longrightarrow \: \: \: \dfrac{4}{1} \: \times \: \dfrac{4}{4} \: = \: \bf{\dfrac{c}{a}}}}⟶ [/tex]
[tex]\begin{gathered} \: \\ \: \large{\sf{\longrightarrow \: \: \: \dfrac{16}{4} \: = \: \bf{\dfrac{c}{a}}}}\end{gathered}
⟶ [/tex]
On comparing, LHS and RHS, we get,
[tex]
\: \: \large{\boxed{\boxed{\tt{a \; = \; 4 \: \: \: and \: \: \: c \; = \; 16}}}} [/tex]
By applying these values in the standard form of quadratic polynomial, we get,
=> p(x) = 4x² + x + 16
[tex]\: \: \large{\underline{\underline{\rm{\leadsto \: \: Thus, \: the \: required \: quadratic \: polynomial \: is \: \boxed{\bf{p(x) \; = \; 4x^{2} \: + \: x \: + \: 16 \: }}}}}} [/tex]
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[tex]\: \: \qquad \qquad \large{\underline{Let's \: know \: more \: :-}}
Let
′[/tex]
• Different types of polynomials are :-
Linear Polynomial
Quadratic Polynomial
Cubic Polynomial
Bi - Quadratic Polynomial