Question

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

Answer 1

Answer ::

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

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 :-

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

$\: \: \large[{\boxed{\boxed{\sf{\alpha \: \times \: \beta \: = \: \bf{\dfrac{c}{a}}}}}}]$

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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,

~ Case I :-

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

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

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 :-

$\: \: \large{\sf{\longrightarrow \: \: \: \alpha \: \times \: \beta \: = \: \bf{\dfrac{c}{a}}}}$

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

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,

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

$\: \\ \: \large{\sf{\longrightarrow \: \: \: \dfrac{16}{4} \: = \: \bf{\dfrac{c}{a}}}}$

On comparing, LHS and RHS, we get,

$\: \: \large{\boxed{\boxed{\tt{a \; = \; 4 \: \: \: and \: \: \: c \; = \; 16}}}}$

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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$\: \: \qquad \qquad \large{\underline{Let's \: know \: more \: :-}}$

• Different types of polynomials are :-

• Linear Polynomial
• Quadratic Polynomial
• Cubic Polynomial
• Bi - Quadratic Polynomial

Answer 2

Given :

• Sum of zeros = -1/4
• Product of zeros = 4

To find :

• Quadratic polynomial

Solution :

Given,

Sum of zeros , α + β = -1/4/

Product of zeros, αβ = 4

As we know that,

Quadratic polynomial = x² - (α + β)x + αβ

Now putting values,

⇒  Quadratic polynomial = x² - (-1/4)x + 4

⇒  Quadratic polynomial = x² + 1/4x + 4

⇒  Quadratic polynomial = x² + x/4 + 4

⇒  Quadratic polynomial : (4x² + x + 16)/4 = 0

⇒  Quadratic polynomial = 4x²+ x + 16

Therefore,

Quadratic polynomial = 4x² + x + 16

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Some informations :

Sum of zeros = -b/a = -coefficient of x/coefficient of x²

Product of zeros = c/a = constant term/coefficient of x²