Question

find the quadratic polynomial whose zeros are 2/3 and -1/4 verify the relation b/w and the zeros of the polynomial

Answer 1

Solution

Given :-

• Zeroes are 2/3 & -1/4

Find :-

• Equation of quadratic polynomial
• relation b/w and the zeros of the polynomial

Explanation

Equation Formula

★ x² - (Sum of zeroes)x + product of zeroes = 0

Let,

given Zeroes are p & q

★ Sum of zeroes = 2/3 + (-1/4)

==> Sum of zeroes = (2*4 - 1*3)/12

==> Sum of zeroes = (8-3)/12

==> Sum of zeroes = 5/12

And,

★ product of zeroes = 2/3 * (-1/4)

==> product of zeroes = -1/6

Required Formula,

==> x² - (5/12)x + (-1/6) = 0

==> 12x² - 5x - 2 = 0

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Relationship B/T coefficient & zeroes

==> Sum of zeroes = -(Coefficient of x)/(Coefficient of x²)

==> Sum of zeroes = -(-5/12)

==> Sum of zeroes = 5/12

And,

==> Product of zeroes = (Constant part )/(Coefficient of x²)

==> Product of zeroes = -(2/12)

==> Product of zeroes = -1/6

Here , relationship are same .

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Answer 2

★ Question ★

→ find the quadratic polynomial whose zeros are 2/3 and -1/4 verify the relation b/w and the zeros of the polynomial.

$\rule{230}2$

◇ sOLUTIOn ◇

Zeroes of the polynomial are $=\frac{2}{3}\:and\:\frac{-1}{4}$

Sum of zeroes =$\longrightarrow \alpha+\beta=\frac{2}{3}+\frac{-1}{4}\\ =\frac{8-3}{12}=\frac{5}{12}$

Product of the zeroes = $\longrightarrow \alpha\beta=\frac{2}{3}×{-1}{4}\\ \frac{-1}{6}$

$\longrightarrow \: General\: equation[x^2-\alpha+\beta+\alpha\beta]\\ \longrightarrow x^2-\frac{5}{12}+\frac{-1}{6}=0\\ \longrightarrow 12x^2-5x-2=0$

Relationship

sum of zeroes =$\frac{-b}{a}\\ \longrightarrow \frac{5}{12}= \frac{5}{12}$ verified

Product of zeroes = $\frac{c}{a}\\ \longrightarrow {-1}{6}=\frac{-2}{12} = \frac{-1}{6}$ verified

Since,

the relation b/w and the zeros of the polynomial is Verified.